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186 Chapter 4 Applications of First Order Equations x y Figure 4.5.10 Curves orthogonal at a point of intersection x y Figure 4.5.11 Orthogonal families of circles and lines are orthogonal trajectories of the integral curves of the differential equation y′ = f(x, y), because at any point (x0, y0) where curves from the two families intersect the slopes of the respective tangent lines are m1 = f(x0, y0) and m2 = − 1 f(x0, y0). This suggests a method for finding orthogonal trajectories of a family of integral curves of a first order equation. Finding Orthogonal Trajectories Step 1. Find a differential equation y′ = f(x, y) for the given family. Step 2. Solve the differential equation y′ = − 1 f(x, y) to find the orthogonal trajectories. Example 4.5.9 Find the orthogonal trajectories of the family of circles x2 + y2 = c2 (c > 0). (4.5.17) Solution To find a differential equation for the family of circles we differentiate (4.5.17) implicitly with respect to x to obtain 2x + 2yy′ = 0, | Elementary Differential Equations with Boundary Value Problems_Page_196_Chunk2601 |
Section 4.5 Applications to Curves 187 or y′ = −x y . Therefore the integral curves of y′ = y x are orthogonal trajectories of the given family. We leave it to you to verify that the general solution of this equation is y = kx, where k is an arbitrary constant. This is the equation of a nonvertical line through (0, 0). The y axis is also an orthogonal trajectory of the given family. Therefore every line through the origin is an orthogonal trajectory of the given family (4.5.17) (Figure 4.5.11). This is consistent with the theorem of plane geometry which states that a diameter of a circle and a tangent line to the circle at the end of the diameter are perpendicular. Example 4.5.10 Find the orthogonal trajectories of the family of hyperbolas xy = c (c ̸= 0) (4.5.18) (Figure 4.5.7). Solution Differentiating (4.5.18) implicitly with respect to x yields y + xy′ = 0, or y′ = −y x; thus, the integral curves of y′ = x y are orthogonal trajectories of the given family. Separating variables yields y′y = x and integrating yields y2 −x2 = k, which is the equation of a hyperbola if k ̸= 0, or of the lines y = x and y = −x if k = 0 (Figure 4.5.12). Example 4.5.11 Find the orthogonal trajectories of the family of circles defined by (x −c)2 + y2 = c2 (c ̸= 0). (4.5.19) These circles are centered on the x-axis and tangent to the y-axis (Figure 4.5.13(a)). Solution Multiplying out the left side of (4.5.19) yields x2 −2cx + y2 = 0, (4.5.20) | Elementary Differential Equations with Boundary Value Problems_Page_197_Chunk2602 |
188 Chapter 4 Applications of First Order Equations x y Figure 4.5.12 Orthogonal trajectories of the hyperbolas xy = c and differentiating this implicitly with respect to x yields 2(x −c) + 2yy′ = 0. (4.5.21) From (4.5.20), c = x2 + y2 2x , so x −c = x −x2 + y2 2x = x2 −y2 2x . Substituting this into (4.5.21) and solving for y′ yields y′ = y2 −x2 2xy . (4.5.22) The curves defined by (4.5.19) are integral curves of (4.5.22), and the integral curves of y′ = 2xy x2 −y2 are orthogonal trajectories of the family (4.5.19). This is a homogeneous nonlinear equation, which we studied in Section 2.4. Substituting y = ux yields u′x + u = 2x(ux) x2 −(ux)2 = 2u 1 −u2 , so u′x = 2u 1 −u2 −u = u(u2 + 1) 1 −u2 , | Elementary Differential Equations with Boundary Value Problems_Page_198_Chunk2603 |
Section 4.5 Applications to Curves 189 Separating variables yields 1 −u2 u(u2 + 1)u′ = 1 x, or, equivalently, 1 u − 2u u2 + 1 u′ = 1 x. Therefore ln |u| −ln(u2 + 1) = ln |x| + k. By substituting u = y/x, we see that ln|y| −ln|x| −ln(x2 + y2) + ln(x2) = ln|x| + k, which, since ln(x2) = 2 ln|x|, is equivalent to ln|y| −ln(x2 + y2) = k, or |y| = ek(x2 + y2). To see what these curves are we rewrite this equation as x2 + |y|2 −e−k|y| = 0 and complete the square to obtain x2 + (|y| −e−k/2)2 = (e−k/2)2. This can be rewritten as x2 + (y −h)2 = h2, where h = e−k 2 if y ≥0, −e−k 2 if y ≤0. Thus, the orthogonal trajectories are circles centered on the y axis and tangent to the x axis (Fig- ure 4.5.13(b)). The circles for which h > 0 are above the x-axis, while those for which h < 0 are below. y y x x (a) (b) Figure 4.5.13 (a) The circles (x −c)2 + y2 = c2 (b) The circles x2 + (y −h)2 = h2 | Elementary Differential Equations with Boundary Value Problems_Page_199_Chunk2604 |
190 Chapter 4 Applications of First Order Equations 4.5 Exercises In Exercises 1–8 find a first order differential equation for the given family of curves. 1. y(x2 + y2) = c 2. exy = cy 3. ln|xy| = c(x2 + y2) 4. y = x1/2 + cx 5. y = ex2 + ce−x2 6. y = x3 + c x 7. y = sin x + cex 8. y = ex + c(1 + x2) 9. Show that the family of circles (x −x0)2 + y2 = 1, −∞< x0 < ∞, can be obtained by joining integral curves of two first order differential equations. More specifi- cally, find differential equations for the families of semicircles (x −x0)2 + y2 = 1, x0 < x < x0 + 1, −∞< x0 < ∞, (x −x0)2 + y2 = 1, x0 −1 < x < x0, −∞< x0 < ∞. 10. Suppose f and g are differentiable for all x. Find a differential equation for the family of functions y = f + cg (c=constant). In Exercises 11–13 find a first order differential equation for the given family of curves. 11. Lines through a given point (x0, y0). 12. Circles through (−1, 0) and (1, 0). 13. Circles through (0, 0) and (0, 2). 14. Use the method Example 4.5.6(a) to find the equations of lines through the given points tangent to the parabola y = x2. Also, find the points of tangency. (a) (5, 9) (b) (6, 11) (c) (−6, 20) (d) (−3, 5) 15. (a) Show that the equation of the line tangent to the circle x2 + y2 = 1 (A) at a point (x0, y0) on the circle is y = 1 −x0x y0 if x0 ̸= ±1. (B) (b) Show that if y′ is the slope of a nonvertical tangent line to the circle (A) and (x, y) is a point on the tangent line then (y′)2(x2 −1) −2xyy′ + y2 −1 = 0. (C) | Elementary Differential Equations with Boundary Value Problems_Page_200_Chunk2605 |
Section 4.5 Applications to Curves 191 (c) Show that the segment of the tangent line (B) on which (x −x0)/y0 > 0 is an integral curve of the differential equation y′ = xy − p x2 + y2 −1 x2 −1 , (D) while the segment on which (x −x0)/y0 < 0 is an integral curve of the differential equation y′ = xy + p x2 + y2 −1 x2 −1 . (E) HINT: Use the quadratic formula to solve (C) for y′. Then substitute (B) for y and choose the ± sign in the quadratic formula so that the resulting expression for y′ reduces to the known slope y′ = −x0/y0. (d) Show that the upper and lower semicircles of (A) are also integral curves of (D) and (E). (e) Find the equations of two lines through (5,5) tangent to the circle (A), and find the points of tangency. 16. (a) Show that the equation of the line tangent to the parabola x = y2 (A) at a point (x0, y0) ̸= (0, 0) on the parabola is y = y0 2 + x 2y0 . (B) (b) Show that if y′ is the slope of a nonvertical tangent line to the parabola (A) and (x, y) is a point on the tangent line then 4x2(y′)2 −4xyy′ + x = 0. (C) (c) Show that the segment of the tangent line defined in (a) on which x > x0 is an integral curve of the differential equation y′ = y + p y2 −x 2x , (D) while the segment on which x < x0 is an integral curve of the differential equation y′ = y − p y2 −x 2x , (E) HINT: Use the quadratic formula to solve (C) for y′. Then substitute (B) for y and choose the ± sign in the quadratic formula so that the resulting expression for y′ reduces to the known slope y′ = 1 2y0 . (d) Show that the upper and lower halves of the parabola (A), given by y = √x and y = −√x for x > 0, are also integral curves of (D) and (E). 17. Use the results of Exercise 16 to find the equations of two lines tangent to the parabola x = y2 and passing through the given point. Also find the points of tangency. (a) (−5, 2) (b) (−4, 0) (c) (7, 4) (d) (5, −3) 18. Find a curve y = y(x) through (1,2) such that the tangent to the curve at any point (x0, y(x0)) intersects the x axis at xI = x0/2. | Elementary Differential Equations with Boundary Value Problems_Page_201_Chunk2606 |
192 Chapter 4 Applications of First Order Equations 19. Find all curves y = y(x) such that the tangent to the curve at any point (x0, y(x0)) intersects the x axis at xI = x3 0. 20. Find all curves y = y(x) such that the tangent to the curve at any point passes through a given point (x1, y1). 21. Find a curve y = y(x) through (1, −1) such that the tangent to the curve at any point (x0, y(x0)) intersects the y axis at yI = x3 0. 22. Find all curves y = y(x) such that the tangent to the curve at any point (x0, y(x0)) intersects the y axis at yI = x0. 23. Find a curve y = y(x) through (0, 2) such that the normal to the curve at any point (x0, y(x0)) intersects the x axis at xI = x0 + 1. 24. Find a curve y = y(x) through (2, 1) such that the normal to the curve at any point (x0, y(x0)) intersects the y axis at yI = 2y(x0). In Exercises 25–29 find the orthogonal trajectories of the given family of curves. 25. x2 + 2y2 = c2 26. x2 + 4xy + y2 = c 27. y = ce2x 28. xyex2 = c 29. y = cex x 30. Find a curve through (−1, 3) orthogonal to every parabola of the form y = 1 + cx2 that it intersects. Which of these parabolas does the desired curve intersect? 31. Show that the orthogonal trajectories of x2 + 2axy + y2 = c satisfy |y −x|a+1|y + x|a−1 = k. 32. If lines L and L1 intersect at (x0, y0) and α is the smallest angle through which L must be rotated counterclockwise about (x0, y0) to bring it into coincidence with L1, we say that α is the angle from L to L1; thus, 0 ≤α < π. If L and L1 are tangents to curves C and C1, respectively, that intersect at (x0, y0), we say that C1 intersects C at the angle α. Use the identity tan(A + B) = tan A + tan B 1 −tan A tan B to show that if C and C1 are intersecting integral curves of y′ = f(x, y) and y′ = f(x, y) + tan α 1 −f(x, y) tan α α ̸= π 2 , respectively, then C1 intersects C at the angle α. 33. Use the result of Exercise 32 to find a family of curves that intersect every nonvertical line through the origin at the angle α = π/4. 34. Use the result of Exercise 32 to find a family of curves that intersect every circle centered at the origin at a given angle α ̸= π/2. | Elementary Differential Equations with Boundary Value Problems_Page_202_Chunk2607 |
CHAPTER 5 Linear Second Order Equations IN THIS CHAPTER we study a particularly important class of second order equations. Because of their many applications in science and engineering, second order differential equation have historically been the most thoroughly studied class of differential equations. Research on the theory of second order differential equations continues to the present day. This chapter is devoted to second order equations that can be written in the form P0(x)y′′ + P1(x)y′ + P2(x)y = F (x). Such equations are said to be linear. As in the case of first order linear equations, (A) is said to be homogeneous if F ≡0, or nonhomogeneous if F ̸≡0. SECTION 5.1 is devoted to the theory of homogeneous linear equations. SECTION 5.2 deals with homogeneous equations of the special form ay′′ + by′ + cy = 0, where a, b, and c are constant (a ̸= 0). When you’ve completed this section you’ll know everything there is to know about solving such equations. SECTION 5.3 presents the theory of nonhomogeneous linear equations. SECTIONS 5.4 AND 5.5 present the method of undetermined coefficients, which can be used to solve nonhomogeneous equations of the form ay′′ + by′ + cy = F (x), where a, b, and c are constants and F has a special form that is still sufficiently general to occur in many applications. In this section we make extensive use of the idea of variation of parameters introduced in Chapter 2. SECTION 5.6 deals with reduction of order, a technique based on the idea of variation of parameters, which enables us to find the general solution of a nonhomogeneous linear second order equation provided that we know one nontrivial (not identically zero) solution of the associated homogeneous equation. SECTION 5.6 deals with the method traditionally called variation of parameters, which enables us to find the general solution of a nonhomogeneous linear second order equation provided that we know two nontrivial solutions (with nonconstant ratio) of the associated homogeneous equation. 193 | Elementary Differential Equations with Boundary Value Problems_Page_203_Chunk2608 |
194 Chapter 5 Linear Second Order Equations 5.1 HOMOGENEOUS LINEAR EQUATIONS A second order differential equation is said to be linear if it can be written as y′′ + p(x)y′ + q(x)y = f(x). (5.1.1) We call the function f on the right a forcing function, since in physical applications it’s often related to a force acting on some system modeled by the differential equation. We say that (5.1.1) is homogeneous if f ≡0 or nonhomogeneous if f ̸≡0. Since these definitions are like the corresponding definitions in Section 2.1 for the linear first order equation y′ + p(x)y = f(x), (5.1.2) it’s natural to expect similarities between methods of solving (5.1.1) and (5.1.2). However, solving (5.1.1) is more difficult than solving (5.1.2). For example, while Theorem 2.1.1 gives a formula for the general solution of (5.1.2) in the case where f ≡0 and Theorem 2.1.2 gives a formula for the case where f ̸≡0, there are no formulas for the general solution of (5.1.1) in either case. Therefore we must be content to solve linear second order equations of special forms. In Section 2.1 we considered the homogeneous equation y′ +p(x)y = 0 first, and then used a nontrivial solution of this equation to find the general solution of the nonhomogeneous equation y′ +p(x)y = f(x). Although the progression from the homogeneous to the nonhomogeneous case isn’t that simple for the linear second order equation, it’s still necessary to solve the homogeneous equation y′′ + p(x)y′ + q(x)y = 0 (5.1.3) in order to solve the nonhomogeneous equation (5.1.1). This section is devoted to (5.1.3). The next theorem gives sufficient conditions for existence and uniqueness of solutions of initial value problems for (5.1.3). We omit the proof. Theorem 5.1.1 Suppose p and q are continuous on an open interval (a, b), let x0 be any point in (a, b), and let k0 and k1 be arbitrary real numbers. Then the initial value problem y′′ + p(x)y′ + q(x)y = 0, y(x0) = k0, y′(x0) = k1 has a unique solution on (a, b). Since y ≡0 is obviously a solution of (5.1.3) we call it the trivial solution. Any other solution is nontrivial. Under the assumptions of Theorem 5.1.1, the only solution of the initial value problem y′′ + p(x)y′ + q(x)y = 0, y(x0) = 0, y′(x0) = 0 on (a, b) is the trivial solution (Exercise 24). The next three examples illustrate concepts that we’ll develop later in this section. You shouldn’t be concerned with how to find the given solutions of the equations in these examples. This will be explained in later sections. Example 5.1.1 The coefficients of y′ and y in y′′ −y = 0 (5.1.4) are the constant functions p ≡0 and q ≡−1, which are continuous on (−∞, ∞). Therefore Theo- rem 5.1.1 implies that every initial value problem for (5.1.4) has a unique solution on (−∞, ∞). | Elementary Differential Equations with Boundary Value Problems_Page_204_Chunk2609 |
Section 5.1 Homogeneous Linear Equations 195 (a) Verify that y1 = ex and y2 = e−x are solutions of (5.1.4) on (−∞, ∞). (b) Verify that if c1 and c2 are arbitrary constants, y = c1ex+c2e−x is a solutionof (5.1.4) on (−∞, ∞). (c) Solve the initial value problem y′′ −y = 0, y(0) = 1, y′(0) = 3. (5.1.5) SOLUTION(a) If y1 = ex then y′ 1 = ex and y′′ 1 = ex = y1, so y′′ 1 −y1 = 0. If y2 = e−x, then y′ 2 = −e−x and y′′ 2 = e−x = y2, so y′′ 2 −y2 = 0. SOLUTION(b) If y = c1ex + c2e−x (5.1.6) then y′ = c1ex −c2e−x (5.1.7) and y′′ = c1ex + c2e−x, so y′′ −y = (c1ex + c2e−x) −(c1ex + c2e−x) = c1(ex −ex) + c2(e−x −e−x) = 0 for all x. Therefore y = c1ex + c2e−x is a solution of (5.1.4) on (−∞, ∞). SOLUTION(c) We can solve (5.1.5) by choosing c1 and c2 in (5.1.6) so that y(0) = 1 and y′(0) = 3. Setting x = 0 in (5.1.6) and (5.1.7) shows that this is equivalent to c1 + c2 = 1 c1 −c2 = 3. Solving these equations yields c1 = 2 and c2 = −1. Therefore y = 2ex −e−x is the unique solution of (5.1.5) on (−∞, ∞). Example 5.1.2 Let ω be a positive constant. The coefficients of y′ and y in y′′ + ω2y = 0 (5.1.8) are the constant functions p ≡0 and q ≡ω2, which are continuous on (−∞, ∞). Therefore Theo- rem 5.1.1 implies that every initial value problem for (5.1.8) has a unique solution on (−∞, ∞). (a) Verify that y1 = cos ωx and y2 = sin ωx are solutions of (5.1.8) on (−∞, ∞). (b) Verify that if c1 and c2 are arbitrary constants then y = c1 cos ωx +c2 sin ωx is a solution of (5.1.8) on (−∞, ∞). (c) Solve the initial value problem y′′ + ω2y = 0, y(0) = 1, y′(0) = 3. (5.1.9) SOLUTION(a) If y1 = cos ωx then y′ 1 = −ω sinωx and y′′ 1 = −ω2 cos ωx = −ω2y1, so y′′ 1 +ω2y1 = 0. If y2 = sin ωx then, y′ 2 = ω cos ωx and y′′ 2 = −ω2 sin ωx = −ω2y2, so y′′ 2 + ω2y2 = 0. | Elementary Differential Equations with Boundary Value Problems_Page_205_Chunk2610 |
196 Chapter 5 Linear Second Order Equations SOLUTION(b) If y = c1 cos ωx + c2 sin ωx (5.1.10) then y′ = ω(−c1 sin ωx + c2 cos ωx) (5.1.11) and y′′ = −ω2(c1 cos ωx + c2 sin ωx), so y′′ + ω2y = −ω2(c1 cos ωx + c2 sin ωx) + ω2(c1 cos ωx + c2 sin ωx) = c1ω2(−cos ωx + cos ωx) + c2ω2(−sin ωx + sin ωx) = 0 for all x. Therefore y = c1 cos ωx + c2 sin ωx is a solution of (5.1.8) on (−∞, ∞). SOLUTION(c) To solve (5.1.9), we must choosing c1 and c2 in (5.1.10) so that y(0) = 1 and y′(0) = 3. Setting x = 0 in (5.1.10) and (5.1.11) shows that c1 = 1 and c2 = 3/ω. Therefore y = cos ωx + 3 ω sin ωx is the unique solution of (5.1.9) on (−∞, ∞). Theorem 5.1.1 implies that if k0 and k1 are arbitrary real numbers then the initial value problem P0(x)y′′ + P1(x)y′ + P2(x)y = 0, y(x0) = k0, y′(x0) = k1 (5.1.12) has a unique solution on an interval (a, b) that contains x0, provided that P0, P1, and P2 are continuous and P0 has no zeros on (a, b). To see this, we rewrite the differential equation in (5.1.12) as y′′ + P1(x) P0(x)y′ + P2(x) P0(x)y = 0 and apply Theorem 5.1.1 with p = P1/P0 and q = P2/P0. Example 5.1.3 The equation x2y′′ + xy′ −4y = 0 (5.1.13) has the form of the differential equation in (5.1.12), with P0(x) = x2, P1(x) = x, and P2(x) = −4, which are are all continuous on (−∞, ∞). However, since P (0) = 0 we must consider solutions of (5.1.13) on (−∞, 0) and (0, ∞). Since P0 has no zeros on these intervals, Theorem 5.1.1 implies that the initial value problem x2y′′ + xy′ −4y = 0, y(x0) = k0, y′(x0) = k1 has a unique solution on (0, ∞) if x0 > 0, or on (−∞, 0) if x0 < 0. (a) Verify that y1 = x2 is a solution of (5.1.13) on (−∞, ∞) and y2 = 1/x2 is a solution of (5.1.13) on (−∞, 0) and (0, ∞). (b) Verify that if c1 and c2 are any constants then y = c1x2 +c2/x2 is a solution of (5.1.13) on (−∞, 0) and (0, ∞). (c) Solve the initial value problem x2y′′ + xy′ −4y = 0, y(1) = 2, y′(1) = 0. (5.1.14) | Elementary Differential Equations with Boundary Value Problems_Page_206_Chunk2611 |
Section 5.1 Homogeneous Linear Equations 197 (d) Solve the initial value problem x2y′′ + xy′ −4y = 0, y(−1) = 2, y′(−1) = 0. (5.1.15) SOLUTION(a) If y1 = x2 then y′ 1 = 2x and y′′ 1 = 2, so x2y′′ 1 + xy′ 1 −4y1 = x2(2) + x(2x) −4x2 = 0 for x in (−∞, ∞). If y2 = 1/x2, then y′ 2 = −2/x3 and y′′ 2 = 6/x4, so x2y′′ 2 + xy′ 2 −4y2 = x2 6 x4 −x 2 x3 −4 x2 = 0 for x in (−∞, 0) or (0, ∞). SOLUTION(b) If y = c1x2 + c2 x2 (5.1.16) then y′ = 2c1x −2c2 x3 (5.1.17) and y′′ = 2c1 + 6c2 x4 , so x2y′′ + xy′ −4y = x2 2c1 + 6c2 x4 + x 2c1x −2c2 x3 −4 c1x2 + c2 x2 = c1(2x2 + 2x2 −4x2) + c2 6 x2 −2 x2 −4 x2 = c1 · 0 + c2 · 0 = 0 for x in (−∞, 0) or (0, ∞). SOLUTION(c) To solve (5.1.14), we choose c1 and c2 in (5.1.16) so that y(1) = 2 and y′(1) = 0. Setting x = 1 in (5.1.16) and (5.1.17) shows that this is equivalent to c1 + c2 = 2 2c1 −2c2 = 0. Solving these equations yields c1 = 1 and c2 = 1. Therefore y = x2 + 1/x2 is the unique solution of (5.1.14) on (0, ∞). SOLUTION(d) We can solve (5.1.15) by choosing c1 and c2 in (5.1.16) so that y(−1) = 2 and y′(−1) = 0. Setting x = −1 in (5.1.16) and (5.1.17) shows that this is equivalent to c1 + c2 = 2 −2c1 + 2c2 = 0. Solving these equations yields c1 = 1 and c2 = 1. Therefore y = x2 + 1/x2 is the unique solution of (5.1.15) on (−∞, 0). | Elementary Differential Equations with Boundary Value Problems_Page_207_Chunk2612 |
198 Chapter 5 Linear Second Order Equations Although the formulas for the solutions of (5.1.14) and (5.1.15) are both y = x2 + 1/x2, you should not conclude that these two initial value problems have the same solution. Remember that a solution of an initial value problem is defined on an interval that contains the initial point; therefore, the solution of (5.1.14) is y = x2 + 1/x2 on the interval (0, ∞), which contains the initial point x0 = 1, while the solution of (5.1.15) is y = x2 + 1/x2 on the interval (−∞, 0), which contains the initial point x0 = −1. The General Solution of a Homogeneous Linear Second Order Equation If y1 and y2 are defined on an interval (a, b) and c1 and c2 are constants, then y = c1y1 + c2y2 is a linear combination of y1 and y2. For example, y = 2 cos x + 7 sin x is a linear combination of y1 = cos x and y2 = sin x, with c1 = 2 and c2 = 7. The next theorem states a fact that we’ve already verified in Examples 5.1.1, 5.1.2, and 5.1.3. Theorem 5.1.2 If y1 and y2 are solutions of the homogeneous equation y′′ + p(x)y′ + q(x)y = 0 (5.1.18) on (a, b), then any linear combination y = c1y1 + c2y2 (5.1.19) of y1 and y2 is also a solution of (5.1.18) on (a, b). Proof If y = c1y1 + c2y2 then y′ = c1y′ 1 + c2y′ 2 and y′′ = c1y′′ 1 + c2y′′ 2 . Therefore y′′ + p(x)y′ + q(x)y = (c1y′′ 1 + c2y′′ 2 ) + p(x)(c1y′ 1 + c2y′ 2) + q(x)(c1y1 + c2y2) = c1 (y′′ 1 + p(x)y′ 1 + q(x)y1) + c2 (y′′ 2 + p(x)y′ 2 + q(x)y2) = c1 · 0 + c2 · 0 = 0, since y1 and y2 are solutions of (5.1.18). We say that {y1, y2} is a fundamental set of solutions of (5.1.18) on (a, b) if every solution of (5.1.18) on (a, b) can be written as a linear combination of y1 and y2 as in (5.1.19). In this case we say that (5.1.19) is general solution of (5.1.18) on (a, b). Linear Independence We need a way to determine whether a given set {y1, y2} of solutions of (5.1.18) is a fundamental set. The next definition will enable us to state necessary and sufficient conditions for this. We say that two functions y1 and y2 defined on an interval (a, b) are linearly independent on (a, b) if neither is a constant multiple of the other on (a, b). (In particular, this means that neither can be the trivial solution of (5.1.18), since, for example, if y1 ≡0 we could write y1 = 0y2.) We’ll also say that the set {y1, y2} is linearly independent on (a, b). Theorem 5.1.3 Suppose p and q are continuous on (a, b). Then a set {y1, y2} of solutions of y′′ + p(x)y′ + q(x)y = 0 (5.1.20) on (a, b) is a fundamental set if and only if {y1, y2} is linearly independent on (a, b). | Elementary Differential Equations with Boundary Value Problems_Page_208_Chunk2613 |
Section 5.1 Homogeneous Linear Equations 199 We’ll present the proof of Theorem 5.1.3 in steps worth regarding as theorems in their own right. However, let’s first interpret Theorem 5.1.3 in terms of Examples 5.1.1, 5.1.2, and 5.1.3. Example 5.1.4 (a) Since ex/e−x = e2x is nonconstant, Theorem 5.1.3 implies that y = c1ex + c2e−x is the general solution of y′′ −y = 0 on (−∞, ∞). (b) Since cos ωx/ sin ωx = cot ωx is nonconstant, Theorem 5.1.3 implies that y = c1 cos ωx + c2 sin ωx is the general solution of y′′ + ω2y = 0 on (−∞, ∞). (c) Since x2/x−2 = x4 is nonconstant, Theorem 5.1.3 implies that y = c1x2 + c2/x2 is the general solution of x2y′′ + xy′ −4y = 0 on (−∞, 0) and (0, ∞). The Wronskian and Abel’s Formula To motivate a result that we need in order to prove Theorem 5.1.3, let’s see what is required to prove that {y1, y2} is a fundamental set of solutions of (5.1.20) on (a, b). Let x0 be an arbitrary point in (a, b), and suppose y is an arbitrary solution of (5.1.20) on (a, b). Then y is the unique solution of the initial value problem y′′ + p(x)y′ + q(x)y = 0, y(x0) = k0, y′(x0) = k1; (5.1.21) that is, k0 and k1 are the numbers obtained by evaluating y and y′ at x0. Moreover, k0 and k1 can be any real numbers, since Theorem 5.1.1 implies that (5.1.21) has a solution no matter how k0 and k1 are chosen. Therefore {y1, y2} is a fundamental set of solutionsof (5.1.20) on (a, b) if and only if it’s possible to write the solution of an arbitrary initial value problem (5.1.21) as y = c1y1 + c2y2. This is equivalent to requiring that the system c1y1(x0) + c2y2(x0) = k0 c1y′ 1(x0) + c2y′ 2(x0) = k1 (5.1.22) has a solution (c1, c2) for every choice of (k0, k1). Let’s try to solve (5.1.22). Multiplying the first equation in (5.1.22) by y′ 2(x0) and the second by y2(x0) yields c1y1(x0)y′ 2(x0) + c2y2(x0)y′ 2(x0) = y′ 2(x0)k0 c1y′ 1(x0)y2(x0) + c2y′ 2(x0)y2(x0) = y2(x0)k1, and subtracting the second equation here from the first yields (y1(x0)y′ 2(x0) −y′ 1(x0)y2(x0)) c1 = y′ 2(x0)k0 −y2(x0)k1. (5.1.23) Multiplying the first equation in (5.1.22) by y′ 1(x0) and the second by y1(x0) yields c1y1(x0)y′ 1(x0) + c2y2(x0)y′ 1(x0) = y′ 1(x0)k0 c1y′ 1(x0)y1(x0) + c2y′ 2(x0)y1(x0) = y1(x0)k1, and subtracting the first equation here from the second yields (y1(x0)y′ 2(x0) −y′ 1(x0)y2(x0)) c2 = y1(x0)k1 −y′ 1(x0)k0. (5.1.24) If y1(x0)y′ 2(x0) −y′ 1(x0)y2(x0) = 0, it’s impossible to satisfy (5.1.23) and (5.1.24) (and therefore (5.1.22)) unless k0 and k1 happen to satisfy y1(x0)k1 −y′ 1(x0)k0 = 0 y′ 2(x0)k0 −y2(x0)k1 = 0. | Elementary Differential Equations with Boundary Value Problems_Page_209_Chunk2614 |
200 Chapter 5 Linear Second Order Equations On the other hand, if y1(x0)y′ 2(x0) −y′ 1(x0)y2(x0) ̸= 0 (5.1.25) we can divide (5.1.23) and (5.1.24) through by the quantity on the left to obtain c1 = y′ 2(x0)k0 −y2(x0)k1 y1(x0)y′ 2(x0) −y′ 1(x0)y2(x0) c2 = y1(x0)k1 −y′ 1(x0)k0 y1(x0)y′ 2(x0) −y′ 1(x0)y2(x0), (5.1.26) no matter how k0 and k1 are chosen. This motivates us to consider conditions on y1 and y2 that imply (5.1.25). Theorem 5.1.4 Suppose p and q are continuous on (a, b), let y1 and y2 be solutions of y′′ + p(x)y′ + q(x)y = 0 (5.1.27) on (a, b), and define W = y1y′ 2 −y′ 1y2. (5.1.28) Let x0 be any point in (a, b). Then W(x) = W(x0)e −R x x0 p(t) dt, a < x < b. (5.1.29) Therefore either W has no zeros in (a, b) or W ≡0 on (a, b). Proof Differentiating (5.1.28) yields W ′ = y′ 1y′ 2 + y1y′′ 2 −y′ 1y′ 2 −y′′ 1 y2 = y1y′′ 2 −y′′ 1 y2. (5.1.30) Since y1 and y2 both satisfy (5.1.27), y′′ 1 = −py′ 1 −qy1 and y′′ 2 = −py′ 2 −qy2. Substituting these into (5.1.30) yields W ′ = −y1 | Elementary Differential Equations with Boundary Value Problems_Page_210_Chunk2615 |
Section 5.1 Homogeneous Linear Equations 201 The expressions in (5.1.26) for c1 and c2 can be written in terms of determinants as c1 = 1 W(x0) k0 y2(x0) k1 y′ 2(x0) and c2 = 1 W(x0) y1(x0) k0 y′ 1(x0) k1 . If you’ve taken linear algebra you may recognize this as Cramer’s rule. Example 5.1.5 Verify Abel’s formula for the following differential equations and the corresponding so- lutions, from Examples 5.1.1, 5.1.2, and 5.1.3: (a) y′′ −y = 0; y1 = ex, y2 = e−x (b) y′′ + ω2y = 0; y1 = cos ωx, y2 = sin ωx (c) x2y′′ + xy′ −4y = 0; y1 = x2, y2 = 1/x2 SOLUTION(a) Since p ≡0, we can verify Abel’s formula by showing that W is constant, which is true, since W(x) = ex e−x ex −e−x = ex(−e−x) −exe−x = −2 for all x. SOLUTION(b) Again, since p ≡0, we can verify Abel’s formula by showing that W is constant, which is true, since W(x) = cos ωx sinωx −ω sin ωx ω cos ωx = cos ωx(ω cos ωx) −(−ω sin ωx) sin ωx = ω(cos2 ωx + sin2 ωx) = ω for all x. SOLUTION(c) Computing the Wronskian of y1 = x2 and y2 = 1/x2 directly yields W = x2 1/x2 2x −2/x3 = x2 −2 x3 −2x 1 x2 = −4 x. (5.1.31) To verify Abel’s formula we rewrite the differential equation as y′′ + 1 xy′ −4 x2 y = 0 to see that p(x) = 1/x. If x0 and x are either both in (−∞, 0) or both in (0, ∞) then Z x x0 p(t) dt = Z x x0 dt t = ln x x0 , so Abel’s formula becomes W(x) = W(x0)e−ln(x/x0) = W(x0)x0 x = − 4 x0 x0 x from (5.1.31) = −4 x, | Elementary Differential Equations with Boundary Value Problems_Page_211_Chunk2616 |
202 Chapter 5 Linear Second Order Equations which is consistent with (5.1.31). The next theorem will enable us to complete the proof of Theorem 5.1.3. Theorem 5.1.5 Suppose p and q are continuous on an open interval (a, b), let y1 and y2 be solutions of y′′ + p(x)y′ + q(x)y = 0 (5.1.32) on (a, b), and let W = y1y′ 2 −y′ 1y2. Then y1 and y2 are linearly independent on (a, b) if and only if W has no zeros on (a, b). Proof We first show that if W(x0) = 0 for some x0 in (a, b), then y1 and y2 are linearly dependent on (a, b). Let I be a subinterval of (a, b) on which y1 has no zeros. (If there’s no such subinterval, y1 ≡0 on (a, b), so y1 and y2 are linearly independent, and we’re finished with this part of the proof.) Then y2/y1 is defined on I, and y2 y1 ′ = y1y′ 2 −y′ 1y2 y2 1 = W y2 1 . (5.1.33) However, if W(x0) = 0, Theorem 5.1.4 implies that W ≡0 on (a, b). Therefore (5.1.33) implies that (y2/y1)′ ≡0, so y2/y1 = c (constant) on I. This shows that y2(x) = cy1(x) for all x in I. However, we want to show that y2 = cy1(x) for all x in (a, b). Let Y = y2 −cy1. Then Y is a solution of (5.1.32) on (a, b) such that Y ≡0 on I, and therefore Y ′ ≡0 on I. Consequently, if x0 is chosen arbitrarily in I then Y is a solution of the initial value problem y′′ + p(x)y′ + q(x)y = 0, y(x0) = 0, y′(x0) = 0, which implies that Y ≡0 on (a, b), by the paragraph following Theorem 5.1.1. (See also Exercise 24). Hence, y2 −cy1 ≡0 on (a, b), which implies that y1 and y2 are not linearly independent on (a, b). Now suppose W has no zeros on (a, b). Then y1 can’t be identically zero on (a, b) (why not?), and therefore there is a subinterval I of (a, b) on which y1 has no zeros. Since (5.1.33) implies that y2/y1 is nonconstant on I, y2 isn’t a constant multiple of y1 on (a, b). A similar argument shows that y1 isn’t a constant multiple of y2 on (a, b), since y1 y2 ′ = y′ 1y2 −y1y′ 2 y2 2 = −W y2 2 on any subinterval of (a, b) where y2 has no zeros. We can now complete the proof of Theorem 5.1.3. From Theorem 5.1.5, two solutions y1 and y2 of (5.1.32) are linearly independent on (a, b) if and only if W has no zeros on (a, b). From Theorem 5.1.4 and the motivating comments preceding it, {y1, y2} is a fundamental set of solutions of (5.1.32) if and only if W has no zeros on (a, b). Therefore {y1, y2} is a fundamental set for (5.1.32) on (a, b) if and only if {y1, y2} is linearly independent on (a, b). The next theorem summarizes the relationships among the concepts discussed in this section. Theorem 5.1.6 Suppose p and q are continuous on an open interval (a, b) and let y1 and y2 be solutions of y′′ + p(x)y′ + q(x)y = 0 (5.1.34) on (a, b). Then the following statements are equivalent; that is, they are either all true or all false. (a) The general solution of (5.1.34) on (a, b) is y = c1y1 + c2y2. (b) {y1, y2} is a fundamental set of solutions of (5.1.34) on (a, b). (c) {y1, y2} is linearly independent on (a, b). | Elementary Differential Equations with Boundary Value Problems_Page_212_Chunk2617 |
Section 5.1 Homogeneous Linear Equations 203 (d) The Wronskian of {y1, y2} is nonzero at some point in (a, b). (e) The Wronskian of {y1, y2} is nonzero at all points in (a, b). We can apply this theorem to an equation written as P0(x)y′′ + P1(x)y′ + P2(x)y = 0 on an interval (a, b) where P0, P1, and P2 are continuous and P0 has no zeros. Theorem 5.1.7 Suppose c is in (a, b) and α and β are real numbers, not both zero. Under the assump- tions of Theorem 5.1.7, suppose y1 and y2 are solutions of (5.1.34) such that αy1(c) + βy′ 1(c) = 0 and αy2(c) + βy′ 2(c) = 0. (5.1.35) Then {y1, y2} isn’t linearly independent on (a, b). Proof Since α and β are not both zero, (5.1.35) implies that y1(c) y′ 1(c) y2(c) y′ 2(c) = 0, so y1(c) y2(c) y′ 1(c) y′ 2(c) = 0 and Theorem 5.1.6 implies the stated conclusion. 5.1 Exercises 1. (a) Verify that y1 = e2x and y2 = e5x are solutions of y′′ −7y′ + 10y = 0 (A) on (−∞, ∞). (b) Verify that if c1 and c2 are arbitrary constants then y = c1e2x + c2e5x is a solution of (A) on (−∞, ∞). (c) Solve the initial value problem y′′ −7y′ + 10y = 0, y(0) = −1, y′(0) = 1. (d) Solve the initial value problem y′′ −7y′ + 10y = 0, y(0) = k0, y′(0) = k1. 2. (a) Verify that y1 = ex cos x and y2 = ex sinx are solutions of y′′ −2y′ + 2y = 0 (A) on (−∞, ∞). (b) Verify that if c1 and c2 are arbitrary constants then y = c1ex cos x + c2ex sin x is a solution of (A) on (−∞, ∞). (c) Solve the initial value problem y′′ −2y′ + 2y = 0, y(0) = 3, y′(0) = −2. | Elementary Differential Equations with Boundary Value Problems_Page_213_Chunk2618 |
204 Chapter 5 Linear Second Order Equations (d) Solve the initial value problem y′′ −2y′ + 2y = 0, y(0) = k0, y′(0) = k1. 3. (a) Verify that y1 = ex and y2 = xex are solutions of y′′ −2y′ + y = 0 (A) on (−∞, ∞). (b) Verify that if c1 and c2 are arbitrary constants then y = ex(c1 + c2x) is a solution of (A) on (−∞, ∞). (c) Solve the initial value problem y′′ −2y′ + y = 0, y(0) = 7, y′(0) = 4. (d) Solve the initial value problem y′′ −2y′ + y = 0, y(0) = k0, y′(0) = k1. 4. (a) Verify that y1 = 1/(x −1) and y2 = 1/(x + 1) are solutions of (x2 −1)y′′ + 4xy′ + 2y = 0 (A) on (−∞, −1), (−1, 1), and (1, ∞). What is the general solution of (A) on each of these intervals? (b) Solve the initial value problem (x2 −1)y′′ + 4xy′ + 2y = 0, y(0) = −5, y′(0) = 1. What is the interval of validity of the solution? (c) C/G Graph the solution of the initial value problem. (d) Verify Abel’s formula for y1 and y2, with x0 = 0. 5. Compute the Wronskians of the given sets of functions. (a) {1, ex} (b) {ex, ex sin x} (c) {x + 1, x2 + 2} (d) {x1/2, x−1/3} (e) {sin x x , cos x x } (f) {x ln|x|, x2 ln |x|} (g) {ex cos √x, ex sin √x} 6. Find the Wronskian of a given set {y1, y2} of solutions of y′′ + 3(x2 + 1)y′ −2y = 0, given that W(π) = 0. 7. Find the Wronskian of a given set {y1, y2} of solutions of (1 −x2)y′′ −2xy′ + α(α + 1)y = 0, given that W(0) = 1. (This is Legendre’s equation.) | Elementary Differential Equations with Boundary Value Problems_Page_214_Chunk2619 |
Section 5.1 Homogeneous Linear Equations 205 8. Find the Wronskian of a given set {y1, y2} of solutions of x2y′′ + xy′ + (x2 −ν2)y = 0, given that W(1) = 1. (This is Bessel’s equation.) 9. (This exercise shows that if you know one nontrivial solution of y′′ + p(x)y′ + q(x)y = 0, you can use Abel’s formula to find another.) Suppose p and q are continuous and y1 is a solution of y′′ + p(x)y′ + q(x)y = 0 (A) that has no zeros on (a, b). Let P (x) = R p(x) dx be any antiderivative of p on (a, b). (a) Show that if K is an arbitrary nonzero constant and y2 satisfies y1y′ 2 −y′ 1y2 = Ke−P(x) (B) on (a, b), then y2 also satisfies (A) on (a, b), and {y1, y2} is a fundamental set of solutions on (A) on (a, b). (b) Conclude from (a) that if y2 = uy1 where u′ = K e−P(x) y2 1(x) , then {y1, y2} is a fundamental set of solutions of (A) on (a, b). In Exercises 10–23 use the method suggested by Exercise 9 to find a second solution y2 that isn’t a constant multiple of the solution y1. Choose K conveniently to simplify y2. 10. y′′ −2y′ −3y = 0; y1 = e3x 11. y′′ −6y′ + 9y = 0; y1 = e3x 12. y′′ −2ay′ + a2y = 0 (a = constant); y1 = eax 13. x2y′′ + xy′ −y = 0; y1 = x 14. x2y′′ −xy′ + y = 0; y1 = x 15. x2y′′ −(2a −1)xy′ + a2y = 0 (a = nonzero constant); x > 0; y1 = xa 16. 4x2y′′ −4xy′ + (3 −16x2)y = 0; y1 = x1/2e2x 17. (x −1)y′′ −xy′ + y = 0; y1 = ex 18. x2y′′ −2xy′ + (x2 + 2)y = 0; y1 = x cos x 19. 4x2(sin x)y′′ −4x(x cos x + sinx)y′ + (2x cos x + 3 sin x)y = 0; y1 = x1/2 20. (3x −1)y′′ −(3x + 2)y′ −(6x −8)y = 0; y1 = e2x 21. (x2 −4)y′′ + 4xy′ + 2y = 0; y1 = 1 x −2 22. (2x + 1)xy′′ −2(2x2 −1)y′ −4(x + 1)y = 0; y1 = 1 x 23. (x2 −2x)y′′ + (2 −x2)y′ + (2x −2)y = 0; y1 = ex 24. Suppose p and q are continuouson an open interval (a, b) and let x0 be in (a, b). Use Theorem 5.1.1 to show that the only solution of the initial value problem y′′ + p(x)y′ + q(x)y = 0, y(x0) = 0, y′(x0) = 0 on (a, b) is the trivial solution y ≡0. | Elementary Differential Equations with Boundary Value Problems_Page_215_Chunk2620 |
206 Chapter 5 Linear Second Order Equations 25. Suppose P0, P1, and P2 are continuous on (a, b) and let x0 be in (a, b). Show that if either of the following statements is true then P0(x) = 0 for some x in (a, b). (a) The initial value problem P0(x)y′′ + P1(x)y′ + P2(x)y = 0, y(x0) = k0, y′(x0) = k1 has more than one solution on (a, b). (b) The initial value problem P0(x)y′′ + P1(x)y′ + P2(x)y = 0, y(x0) = 0, y′(x0) = 0 has a nontrivial solution on (a, b). 26. Suppose p and q are continuous on (a, b) and y1 and y2 are solutions of y′′ + p(x)y′ + q(x)y = 0 (A) on (a, b). Let z1 = αy1 + βy2 and z2 = γy1 + δy2, where α, β, γ, and δ are constants. Show that if {z1, z2} is a fundamental set of solutions of (A) on (a, b) then so is {y1, y2}. 27. Suppose p and q are continuous on (a, b) and {y1, y2} is a fundamental set of solutions of y′′ + p(x)y′ + q(x)y = 0 (A) on (a, b). Let z1 = αy1 + βy2 and z2 = γy1 + δy2, where α, β, γ, and δ are constants. Show that {z1, z2} is a fundamental set of solutions of (A) on (a, b) if and only if αγ −βδ ̸= 0. 28. Suppose y1 is differentiable on an interval (a, b) and y2 = ky1, where k is a constant. Show that the Wronskian of {y1, y2} is identically zero on (a, b). 29. Let y1 = x3 and y2 = x3, x ≥0, −x3, x < 0. (a) Show that the Wronskian of {y1, y2} is defined and identically zero on (−∞, ∞). (b) Suppose a < 0 < b. Show that {y1, y2} is linearly independent on (a, b). (c) Use Exercise 25(b) to show that these results don’t contradict Theorem 5.1.5, because neither y1 nor y2 can be a solution of an equation y′′ + p(x)y′ + q(x)y = 0 on (a, b) if p and q are continuous on (a, b). 30. Suppose p and q are continuous on (a, b) and {y1, y2} is a set of solutions of y′′ + p(x)y′ + q(x)y = 0 on (a, b) such that either y1(x0) = y2(x0) = 0 or y′ 1(x0) = y′ 2(x0) = 0 for some x0 in (a, b). Show that {y1, y2} is linearly dependent on (a, b). | Elementary Differential Equations with Boundary Value Problems_Page_216_Chunk2621 |
Section 5.1 Homogeneous Linear Equations 207 31. Suppose p and q are continuous on (a, b) and {y1, y2} is a fundamental set of solutions of y′′ + p(x)y′ + q(x)y = 0 on (a, b). Show that if y1(x1) = y1(x2) = 0, where a < x1 < x2 < b, then y2(x) = 0 for some x in (x1, x2). HINT: Show that if y2 has no zeros in (x1, x2), then y1/y2 is either strictly increasing or strictly decreasing on (x1, x2), and deduce a contradiction. 32. Suppose p and q are continuous on (a, b) and every solution of y′′ + p(x)y′ + q(x)y = 0 (A) on (a, b) can be written as a linear combination of the twice differentiable functions {y1, y2}. Use Theorem 5.1.1 to show that y1 and y2 are themselves solutions of (A) on (a, b). 33. Suppose p1, p2, q1, and q2 are continuous on (a, b) and the equations y′′ + p1(x)y′ + q1(x)y = 0 and y′′ + p2(x)y′ + q2(x)y = 0 have the same solutions on (a, b). Show that p1 = p2 and q1 = q2 on (a, b). HINT: Use Abel’s formula. 34. (For this exercise you have to know about 3 × 3 determinants.) Show that if y1 and y2 are twice continuously differentiable on (a, b) and the Wronskian W of {y1, y2} has no zeros in (a, b) then the equation 1 W y y1 y2 y′ y′ 1 y′ 2 y′′ y′′ 1 y′′ 2 = 0 can be written as y′′ + p(x)y′ + q(x)y = 0, (A) where p and q are continuous on (a, b) and {y1, y2} is a fundamental set of solutions of (A) on (a, b). HINT: Expand the determinant by cofactors of its first column. 35. Use the method suggested by Exercise 34 to find a linear homogeneous equation for which the given functions form a fundamental set of solutions on some interval. (a) ex cos 2x, ex sin2x (b) x, e2x (c) x, x lnx (d) cos(lnx), sin(ln x) (e) cosh x, sinh x (f) x2 −1, x2 + 1 36. Suppose p and q are continuous on (a, b) and {y1, y2} is a fundamental set of solutions of y′′ + p(x)y′ + q(x)y = 0 (A) on (a, b). Show that if y is a solution of (A) on (a, b), there’s exactly one way to choose c1 and c2 so that y = c1y1 + c2y2 on (a, b). 37. Suppose p and q are continuous on (a, b) and x0 is in (a, b). Let y1 and y2 be the solutions of y′′ + p(x)y′ + q(x)y = 0 (A) such that y1(x0) = 1, y′ 1(x0) = 0 and y2(x0) = 0, y′ 2(x0) = 1. (Theorem 5.1.1 implies that each of these initial value problems has a unique solution on (a, b).) | Elementary Differential Equations with Boundary Value Problems_Page_217_Chunk2622 |
208 Chapter 5 Linear Second Order Equations (a) Show that {y1, y2} is linearly independent on (a, b). (b) Show that an arbitrary solution y of (A) on (a, b) can be written as y = y(x0)y1 + y′(x0)y2. (c) Express the solution of the initial value problem y′′ + p(x)y′ + q(x)y = 0, y(x0) = k0, y′(x0) = k1 as a linear combination of y1 and y2. 38. Find solutions y1 and y2 of the equation y′′ = 0 that satisfy the initial conditions y1(x0) = 1, y′ 1(x0) = 0 and y2(x0) = 0, y′ 2(x0) = 1. Then use Exercise 37 (c) to write the solution of the initial value problem y′′ = 0, y(0) = k0, y′(0) = k1 as a linear combination of y1 and y2. 39. Let x0 be an arbitrary real number. Given (Example 5.1.1) that ex and e−x are solutionsof y′′−y = 0, find solutions y1 and y2 of y′′ −y = 0 such that y1(x0) = 1, y′ 1(x0) = 0 and y2(x0) = 0, y′ 2(x0) = 1. Then use Exercise 37 (c) to write the solution of the initial value problem y′′ −y = 0, y(x0) = k0, y′(x0) = k1 as a linear combination of y1 and y2. 40. Let x0 be an arbitrary real number. Given (Example 5.1.2) that cos ωx and sin ωx are solutions of y′′ + ω2y = 0, find solutions of y′′ + ω2y = 0 such that y1(x0) = 1, y′ 1(x0) = 0 and y2(x0) = 0, y′ 2(x0) = 1. Then use Exercise 37 (c) to write the solution of the initial value problem y′′ + ω2y = 0, y(x0) = k0, y′(x0) = k1 as a linear combination of y1 and y2. Use the identities cos(A + B) = cos A cos B −sin A sin B sin(A + B) = sin A cos B + cos A sin B to simplify your expressions for y1, y2, and y. 41. Recall from Exercise 4 that 1/(x −1) and 1/(x + 1) are solutions of (x2 −1)y′′ + 4xy′ + 2y = 0 (A) on (−1, 1). Find solutions of (A) such that y1(0) = 1, y′ 1(0) = 0 and y2(0) = 0, y′ 2(0) = 1. Then use Exercise 37 (c) to write the solution of initial value problem (x2 −1)y′′ + 4xy′ + 2y = 0, y(0) = k0, y′(0) = k1 as a linear combination of y1 and y2. | Elementary Differential Equations with Boundary Value Problems_Page_218_Chunk2623 |
Section 5.1 Homogeneous Linear Equations 209 42. (a) Verify that y1 = x2 and y2 = x3 satisfy x2y′′ −4xy′ + 6y = 0 (A) on (−∞, ∞) and that {y1, y2} is a fundamental set of solutions of (A) on (−∞, 0) and (0, ∞). (b) Let a1, a2, b1, and b2 be constants. Show that y = a1x2 + a2x3, x ≥0, b1x2 + b2x3, x < 0 is a solution of (A) on (−∞, ∞) if and only if a1 = b1. From this, justify the statement that y is a solution of (A) on (−∞, ∞) if and only if y = c1x2 + c2x3, x ≥0, c1x2 + c3x3, x < 0, where c1, c2, and c3 are arbitrary constants. (c) For what values of k0 and k1 does the initial value problem x2y′′ −4xy′ + 6y = 0, y(0) = k0, y′(0) = k1 have a solution? What are the solutions? (d) Show that if x0 ̸= 0 and k0, k1 are arbitrary constants, the initial value problem x2y′′ −4xy′ + 6y = 0, y(x0) = k0, y′(x0) = k1 (B) has infinitely many solutionson (−∞, ∞). On what interval does (B) have a unique solution? 43. (a) Verify that y1 = x and y2 = x2 satisfy x2y′′ −2xy′ + 2y = 0 (A) on (−∞, ∞) and that {y1, y2} is a fundamental set of solutions of (A) on (−∞, 0) and (0, ∞). (b) Let a1, a2, b1, and b2 be constants. Show that y = a1x + a2x2, x ≥0, b1x + b2x2, x < 0 is a solution of (A) on (−∞, ∞) if and only if a1 = b1 and a2 = b2. From this, justify the statement that the general solution of (A) on (−∞, ∞) is y = c1x + c2x2, where c1 and c2 are arbitrary constants. (c) For what values of k0 and k1 does the initial value problem x2y′′ −2xy′ + 2y = 0, y(0) = k0, y′(0) = k1 have a solution? What are the solutions? (d) Show that if x0 ̸= 0 and k0, k1 are arbitrary constants then the initial value problem x2y′′ −2xy′ + 2y = 0, y(x0) = k0, y′(x0) = k1 has a unique solution on (−∞, ∞). | Elementary Differential Equations with Boundary Value Problems_Page_219_Chunk2624 |
210 Chapter 5 Linear Second Order Equations 44. (a) Verify that y1 = x3 and y2 = x4 satisfy x2y′′ −6xy′ + 12y = 0 (A) on (−∞, ∞), and that {y1, y2} is a fundamental set of solutions of (A) on (−∞, 0) and (0, ∞). (b) Show that y is a solution of (A) on (−∞, ∞) if and only if y = a1x3 + a2x4, x ≥0, b1x3 + b2x4, x < 0, where a1, a2, b1, and b2 are arbitrary constants. (c) For what values of k0 and k1 does the initial value problem x2y′′ −6xy′ + 12y = 0, y(0) = k0, y′(0) = k1 have a solution? What are the solutions? (d) Show that if x0 ̸= 0 and k0, k1 are arbitrary constants then the initial value problem x2y′′ −6xy′ + 12y = 0, y(x0) = k0, y′(x0) = k1 (B) has infinitely many solutionson (−∞, ∞). On what interval does (B) have a unique solution? 5.2 CONSTANT COEFFICIENT HOMOGENEOUS EQUATIONS If a, b, and c are real constants and a ̸= 0, then ay′′ + by′ + cy = F (x) is said to be a constant coefficient equation. In this section we consider the homogeneous constant coef- ficient equation ay′′ + by′ + cy = 0. (5.2.1) As we’ll see, all solutions of (5.2.1) are defined on (−∞, ∞). This being the case, we’ll omit references to the interval on which solutions are defined, or on which a given set of solutions is a fundamental set, etc., since the interval will always be (−∞, ∞). The key to solving (5.2.1) is that if y = erx where r is a constant then the left side of (5.2.1) is a multiple of erx; thus, if y = erx then y′ = rerx and y′′ = r2erx, so ay′′ + by′ + cy = ar2erx + brerx + cerx = (ar2 + br + c)erx. (5.2.2) The quadratic polynomial p(r) = ar2 + br + c is the characteristic polynomial of (5.2.1), and p(r) = 0 is the characteristic equation. From (5.2.2) we can see that y = erx is a solution of (5.2.1) if and only if p(r) = 0. The roots of the characteristic equation are given by the quadratic formula r = −b ± √ b2 −4ac 2a . (5.2.3) We consider three cases: | Elementary Differential Equations with Boundary Value Problems_Page_220_Chunk2625 |
Section 5.2 Constant Coefficient Homogeneous Equations 211 CASE 1. b2 −4ac > 0, so the characteristic equation has two distinct real roots. CASE 2. b2 −4ac = 0, so the characteristic equation has a repeated real root. CASE 3. b2 −4ac < 0, so the characteristic equation has complex roots. In each case we’ll start with an example. Case 1: Distinct Real Roots Example 5.2.1 (a) Find the general solution of y′′ + 6y′ + 5y = 0. (5.2.4) (b) Solve the initial value problem y′′ + 6y′ + 5y = 0, y(0) = 3, y′(0) = −1. (5.2.5) SOLUTION(a) The characteristic polynomial of (5.2.4) is p(r) = r2 + 6r + 5 = (r + 1)(r + 5). Since p(−1) = p(−5) = 0, y1 = e−x and y2 = e−5x are solutions of (5.2.4). Since y2/y1 = e−4x is nonconstant, 5.1.6 implies that the general solution of (5.2.4) is y = c1e−x + c2e−5x. (5.2.6) SOLUTION(b) We must determine c1 and c2 in (5.2.6) so that y satisfies the initial conditions in (5.2.5). Differentiating (5.2.6) yields y′ = −c1e−x −5c2e−5x. (5.2.7) Imposing the initial conditions y(0) = 3, y′(0) = −1 in (5.2.6) and (5.2.7) yields c1 + c2 = 3 −c1 −5c2 = −1. The solution of this system is c1 = 7/2, c2 = −1/2. Therefore the solution of (5.2.5) is y = 7 2e−x −1 2e−5x. Figure 5.2.1 is a graph of this solution. If the characteristic equation has arbitrary distinct real roots r1 and r2, then y1 = er1x and y2 = er2x are solutions of ay′′ + by′ + cy = 0. Since y2/y1 = e(r2−r1)x is nonconstant, Theorem 5.1.6 implies that {y1, y2} is a fundamental set of solutions of ay′′ + by′ + cy = 0. | Elementary Differential Equations with Boundary Value Problems_Page_221_Chunk2626 |
212 Chapter 5 Linear Second Order Equations 1 2 3 4 3 2 1 5 x y Figure 5.2.1 y = 7 2e−x −1 2e−5x Case 2: A Repeated Real Root Example 5.2.2 (a) Find the general solution of y′′ + 6y′ + 9y = 0. (5.2.8) (b) Solve the initial value problem y′′ + 6y′ + 9y = 0, y(0) = 3, y′(0) = −1. (5.2.9) SOLUTION(a) The characteristic polynomial of (5.2.8) is p(r) = r2 + 6r + 9 = (r + 3)2, so the characteristic equation has the repeated real root r1 = −3. Therefore y1 = e−3x is a solution of (5.2.8). Since the characteristic equation has no other roots, (5.2.8) has no other solutions of the form erx. We look for solutions of the form y = uy1 = ue−3x, where u is a function that we’ll now determine. (This should remind you of the method of variation of parameters used in Section 2.1 to solve the nonhomogeneous equation y′ + p(x)y = f(x), given a solution y1 of the complementary equation y′ + p(x)y = 0. It’s also a special case of a method called reduction of order that we’ll study in Section 5.6. For other ways to obtain a second solution of (5.2.8) that’s not a multiple of e−3x, see Exercises 5.1.9, 5.1.12, and 33. | Elementary Differential Equations with Boundary Value Problems_Page_222_Chunk2627 |
Section 5.2 Constant Coefficient Homogeneous Equations 213 If y = ue−3x, then y′ = u′e−3x −3ue−3x and y′′ = u′′e−3x −6u′e−3x + 9ue−3x, so y′′ + 6y′ + 9y = e−3x [(u′′ −6u′ + 9u) + 6(u′ −3u) + 9u] = e−3x [u′′ −(6 −6)u′ + (9 −18 + 9)u] = u′′e−3x. Therefore y = ue−3x is a solution of (5.2.8) if and only if u′′ = 0, which is equivalent to u = c1 + c2x, where c1 and c2 are constants. Therefore any function of the form y = e−3x(c1 + c2x) (5.2.10) is a solution of (5.2.8). Letting c1 = 1 and c2 = 0 yields the solution y1 = e−3x that we already knew. Letting c1 = 0 and c2 = 1 yields the second solution y2 = xe−3x. Since y2/y1 = x is nonconstant, 5.1.6 implies that {y1, y2} is fundamental set of solutions of (5.2.8), and (5.2.10) is the general solution. SOLUTION(b) Differentiating (5.2.10) yields y′ = −3e−3x(c1 + c2x) + c2e−3x. (5.2.11) Imposing the initial conditions y(0) = 3, y′(0) = −1 in (5.2.10) and (5.2.11) yields c1 = 3 and −3c1 + c2 = −1, so c2 = 8. Therefore the solution of (5.2.9) is y = e−3x(3 + 8x). Figure 5.2.2 is a graph of this solution. 1 2 3 1 2 3 x y Figure 5.2.2 y = e−3x(3 + 8x) | Elementary Differential Equations with Boundary Value Problems_Page_223_Chunk2628 |
214 Chapter 5 Linear Second Order Equations If the characteristic equation of ay′′ + by′ + cy = 0 has an arbitrary repeated root r1, the characteristic polynomial must be p(r) = a(r −r1)2 = a(r2 −2r1r + r2 1). Therefore ar2 + br + c = ar2 −(2ar1)r + ar2 1, which implies that b = −2ar1 and c = ar2 1. Therefore ay′′ + by′ + cy = 0 can be written as a(y′′ − 2r1y′ + r2 1y) = 0. Since a ̸= 0 this equation has the same solutions as y′′ −2r1y′ + r2 1y = 0. (5.2.12) Since p(r1) = 0, t y1 = er1x is a solution of ay′′ +by′ +cy = 0, and therefore of (5.2.12). Proceeding as in Example 5.2.2, we look for other solutions of (5.2.12) of the form y = uer1x; then y′ = u′er1x + ruer1x and y′′ = u′′er1x + 2r1u′er1x + r2 1uer1x, so y′′ −2r1y′ + r2 1y = erx (u′′ + 2r1u′ + r2 1u) −2r1(u′ + r1u) + r2 1u = er1x u′′ + (2r1 −2r1)u′ + (r2 1 −2r2 1 + r2 1)u = u′′er1x. Therefore y = uer1x is a solution of (5.2.12) if and only if u′′ = 0, which is equivalent to u = c1 + c2x, where c1 and c2 are constants. Hence, any function of the form y = er1x(c1 + c2x) (5.2.13) is a solution of (5.2.12). Letting c1 = 1 and c2 = 0 here yields the solution y1 = er1x that we already knew. Letting c1 = 0 and c2 = 1 yields the second solution y2 = xer1x. Since y2/y1 = x is noncon- stant, 5.1.6 implies that {y1, y2} is a fundamental set of solutions of (5.2.12), and (5.2.13) is the general solution. Case 3: Complex Conjugate Roots Example 5.2.3 (a) Find the general solution of y′′ + 4y′ + 13y = 0. (5.2.14) (b) Solve the initial value problem y′′ + 4y′ + 13y = 0, y(0) = 2, y′(0) = −3. (5.2.15) SOLUTION(a) The characteristic polynomial of (5.2.14) is p(r) = r2 + 4r + 13 = r2 + 4r + 4 + 9 = (r + 2)2 + 9. The roots of the characteristic equation are r1 = −2 + 3i and r2 = −2 −3i. By analogy with Case 1, it’s reasonable to expect that e(−2+3i)x and e(−2−3i)x are solutions of (5.2.14). This is true (see Exercise 34); however, there are difficulties here, since you are probably not familiar with exponential functions with complex arguments, and even if you are, it’s inconvenient to work with them, since they are complex– valued. We’ll take a simpler approach, which we motivate as follows: the exponential notation suggests that e(−2+3i)x = e−2xe3ix and e(−2−3i)x = e−2xe−3ix, | Elementary Differential Equations with Boundary Value Problems_Page_224_Chunk2629 |
Section 5.2 Constant Coefficient Homogeneous Equations 215 so even though we haven’t defined e3ix and e−3ix, it’s reasonable to expect that every linear combination of e(−2+3i)x and e(−2−3i)x can be written as y = ue−2x, where u depends upon x. To determine u, we note that if y = ue−2x then y′ = u′e−2x −2ue−2x and y′′ = u′′e−2x −4u′e−2x + 4ue−2x, so y′′ + 4y′ + 13y = e−2x [(u′′ −4u′ + 4u) + 4(u′ −2u) + 13u] = e−2x [u′′ −(4 −4)u′ + (4 −8 + 13)u] = e−2x(u′′ + 9u). Therefore y = ue−2x is a solution of (5.2.14) if and only if u′′ + 9u = 0. From Example 5.1.2, the general solution of this equation is u = c1 cos 3x + c2 sin 3x. Therefore any function of the form y = e−2x(c1 cos 3x + c2 sin3x) (5.2.16) is a solution of (5.2.14). Letting c1 = 1 and c2 = 0 yields the solution y1 = e−2x cos 3x. Letting c1 = 0 and c2 = 1 yields the second solution y2 = e−2x sin 3x. Since y2/y1 = tan 3x is nonconstant, 5.1.6 implies that {y1, y2} is a fundamental set of solutions of (5.2.14), and (5.2.16) is the general solution. SOLUTION(b) Imposing the condition y(0) = 2 in (5.2.16) shows that c1 = 2. Differentiating (5.2.16) yields y′ = −2e−2x(c1 cos 3x + c2 sin 3x) + 3e−2x(−c1 sin 3x + c2 cos 3x), and imposing the initial condition y′(0) = −3 here yields −3 = −2c1 + 3c2 = −4 + 3c2, so c2 = 1/3. Therefore the solution of (5.2.15) is y = e−2x(2 cos 3x + 1 3 sin 3x). Figure 5.2.3 is a graph of this function. Now suppose the characteristic equation of ay′′ + by′ + cy = 0 has arbitrary complex roots; thus, b2 −4ac < 0 and, from (5.2.3), the roots are r1 = −b + i √ 4ac −b2 2a , r2 = −b −i √ 4ac −b2 2a , which we rewrite as r1 = λ + iω, r2 = λ −iω, (5.2.17) with λ = −b 2a, ω = √ 4ac −b2 2a . Don’t memorize these formulas. Just remember that r1 and r2 are of the form (5.2.17), where λ is an arbitrary real number and ω is positive; λ and ω are the real and imaginary parts, respectively, of r1. Similarly, λ and −ω are the real and imaginary parts of r2. We say that r1 and r2 are complex conjugates, | Elementary Differential Equations with Boundary Value Problems_Page_225_Chunk2630 |
216 Chapter 5 Linear Second Order Equations 1 2 1 2 x y Figure 5.2.3 y = e−2x(2 cos 3x + 1 3 sin 3x) which means that they have the same real part and their imaginary parts have the same absolute values, but opposite signs. As in Example 5.2.3, it’s reasonable to to expect that the solutions of ay′′ + by′ + cy = 0 are linear combinations of e(λ+iω)x and e(λ−iω)x. Again, the exponential notation suggests that e(λ+iω)x = eλxeiωx and e(λ−iω)x = eλxe−iωx, so even though we haven’t defined eiωx and e−iωx, it’s reasonable to expect that every linear combination of e(λ+iω)x and e(λ−iω)x can be written as y = ueλx, where u depends upon x. To determine u we first observe that since r1 = λ + iω and r2 = λ −iω are the roots of the characteristic equation, p must be of the form p(r) = a(r −r1)(r −r2) = a(r −λ −iω)(r −λ + iω) = a (r −λ)2 + ω2 = a(r2 −2λr + λ2 + ω2). Therefore ay′′ + by′ + cy = 0 can be written as a y′′ −2λy′ + (λ2 + ω2)y = 0. Since a ̸= 0 this equation has the same solutions as y′′ −2λy′ + (λ2 + ω2)y = 0. (5.2.18) To determine u we note that if y = ueλx then y′ = u′eλx + λueλx and y′′ = u′′eλx + 2λu′eλx + λ2ueλx. | Elementary Differential Equations with Boundary Value Problems_Page_226_Chunk2631 |
Section 5.2 Constant Coefficient Homogeneous Equations 217 Substituting these expressions into (5.2.18) and dropping the common factor eλx yields (u′′ + 2λu′ + λ2u) −2λ(u′ + λu) + (λ2 + ω2)u = 0, which simplifies to u′′ + ω2u = 0. From Example 5.1.2, the general solution of this equation is u = c1 cos ωx + c2 sin ωx. Therefore any function of the form y = eλx(c1 cos ωx + c2 sin ωx) (5.2.19) is a solution of (5.2.18). Letting c1 = 1 and c2 = 0 here yields the solution y1 = eλx cos ωx. Letting c1 = 0 and c2 = 1 yields a second solution y2 = eλx sin ωx. Since y2/y1 = tan ωx is nonconstant, so Theorem 5.1.6 implies that {y1, y2} is a fundamental set of solutions of (5.2.18), and (5.2.19) is the general solution. Summary The next theorem summarizes the results of this section. Theorem 5.2.1 Let p(r) = ar2 + br + c be the characteristic polynomial of ay′′ + by′ + cy = 0. (5.2.20) Then: (a) If p(r) = 0 has distinct real roots r1 and r2, then the general solution of (5.2.20) is y = c1er1x + c2er2x. (b) If p(r) = 0 has a repeated root r1, then the general solution of (5.2.20) is y = er1x(c1 + c2x). (c) If p(r) = 0 has complex conjugate roots r1 = λ + iω and r2 = λ −iω (where ω > 0), then the general solution of (5.2.20) is y = eλx(c1 cos ωx + c2 sin ωx). 5.2 Exercises In Exercises 1–12 find the general solution. 1. y′′ + 5y′ −6y = 0 2. y′′ −4y′ + 5y = 0 3. y′′ + 8y′ + 7y = 0 4. y′′ −4y′ + 4y = 0 5. y′′ + 2y′ + 10y = 0 6. y′′ + 6y′ + 10y = 0 | Elementary Differential Equations with Boundary Value Problems_Page_227_Chunk2632 |
218 Chapter 5 Linear Second Order Equations 7. y′′ −8y′ + 16y = 0 8. y′′ + y′ = 0 9. y′′ −2y′ + 3y = 0 10. y′′ + 6y′ + 13y = 0 11. 4y′′ + 4y′ + 10y = 0 12. 10y′′ −3y′ −y = 0 In Exercises 13–17 solve the initial value problem. 13. y′′ + 14y′ + 50y = 0, y(0) = 2, y′(0) = −17 14. 6y′′ −y′ −y = 0, y(0) = 10, y′(0) = 0 15. 6y′′ + y′ −y = 0, y(0) = −1, y′(0) = 3 16. 4y′′ −4y′ −3y = 0, y(0) = 13 12, y′(0) = 23 24 17. 4y′′ −12y′ + 9y = 0, y(0) = 3, y′(0) = 5 2 In Exercises 18–21 solve the initial value problem and graph the solution. 18. C/G y′′ + 7y′ + 12y = 0, y(0) = −1, y′(0) = 0 19. C/G y′′ −6y′ + 9y = 0, y(0) = 0, y′(0) = 2 20. C/G 36y′′ −12y′ + y = 0, y(0) = 3, y′(0) = 5 2 21. C/G y′′ + 4y′ + 10y = 0, y(0) = 3, y′(0) = −2 22. (a) Suppose y is a solution of the constant coefficient homogeneous equation ay′′ + by′ + cy = 0. (A) Let z(x) = y(x −x0), where x0 is an arbitrary real number. Show that az′′ + bz′ + cz = 0. (b) Let z1(x) = y1(x −x0) and z2(x) = y2(x −x0), where {y1, y2} is a fundamental set of solutions of (A). Show that {z1, z2} is also a fundamental set of solutions of (A). (c) The statement of Theorem 5.2.1 is convenient for solving an initial value problem ay′′ + by′ + cy = 0, y(0) = k0, y′(0) = k1, where the initial conditions are imposed at x0 = 0. However, if the initial value problem is ay′′ + by′ + cy = 0, y(x0) = k0, y′(x0) = k1, (B) where x0 ̸= 0, then determining the constants in y = c1er1x + c2er2x, y = er1x(c1 + c2x), or y = eλx(c1 cos ωx + c2 sin ωx) (whichever is applicable) is more complicated. Use (b) to restate Theorem 5.2.1 in a form more convenient for solving (B). In Exercises 23–28 use a method suggested by Exercise 22 to solve the initial value problem. 23. y′′ + 3y′ + 2y = 0, y(1) = −1, y′(1) = 4 | Elementary Differential Equations with Boundary Value Problems_Page_228_Chunk2633 |
Section 5.2 Constant Coefficient Homogeneous Equations 219 24. y′′ −6y′ −7y = 0, y(2) = −1 3, y′(2) = −5 25. y′′ −14y′ + 49y = 0, y(1) = 2, y′(1) = 11 26. 9y′′ + 6y′ + y = 0, y(2) = 2, y′(2) = −14 3 27. 9y′′ + 4y = 0, y(π/4) = 2, y′(π/4) = −2 28. y′′ + 3y = 0, y(π/3) = 2, y′(π/3) = −1 29. Prove: If the characteristic equation of ay′′ + by′ + cy = 0 (A) has a repeated negative root or two roots with negative real parts, then every solution of (A) ap- proaches zero as x →∞. 30. Suppose the characteristic polynomial of ay′′ + by′ + cy = 0 has distinct real roots r1 and r2. Use a method suggested by Exercise 22 to find a formula for the solution of ay′′ + by′ + cy = 0, y(x0) = k0, y′(x0) = k1. 31. Suppose the characteristic polynomial of ay′′ + by′ + cy = 0 has a repeated real root r1. Use a method suggested by Exercise 22 to find a formula for the solution of ay′′ + by′ + cy = 0, y(x0) = k0, y′(x0) = k1. 32. Suppose the characteristic polynomial of ay′′ + by′ + cy = 0 has complex conjugate roots λ ± iω. Use a method suggested by Exercise 22 to find a formula for the solution of ay′′ + by′ + cy = 0, y(x0) = k0, y′(x0) = k1. 33. Suppose the characteristic equation of ay′′ + by′ + cy = 0 (A) has a repeated real root r1. Temporarily, think of erx as a function of two real variables x and r. (a) Show that a ∂2 ∂2x(erx) + b ∂ ∂x(erx) + cerx = a(r −r1)2erx. (B) (b) Differentiate (B) with respect to r to obtain a ∂ ∂r ∂2 ∂2x(erx) + b ∂ ∂r ∂ ∂x(erx) + c(xerx) = [2 + (r −r1)x]a(r −r1)erx. (C) (c) Reverse the orders of the partial differentiations in the first two terms on the left side of (C) to obtain a ∂2 ∂x2 (xerx) + b ∂ ∂x(xerx) + c(xerx) = [2 + (r −r1)x]a(r −r1)erx. (D) (d) Set r = r1 in (B) and (D) to see that y1 = er1x and y2 = xer1x are solutions of (A) | Elementary Differential Equations with Boundary Value Problems_Page_229_Chunk2634 |
220 Chapter 5 Linear Second Order Equations 34. In calculus you learned that eu, cos u, and sin u can be represented by the infinite series eu = ∞ X n=0 un n! = 1 + u 1! + u2 2! + u3 3! + · · · + un n! + · · · (A) cos u = ∞ X n=0 (−1)n u2n (2n)! = 1 −u2 2! + u4 4! + · · · + (−1)n u2n (2n)! + · · · , (B) and sin u = ∞ X n=0 (−1)n u2n+1 (2n + 1)! = u −u3 3! + u5 5! + · · · + (−1)n u2n+1 (2n + 1)! + · · · (C) for all real values of u. Even though you have previously considered (A) only for real values of u, we can set u = iθ, where θ is real, to obtain eiθ = ∞ X n=0 (iθ)n n! . (D) Given the proper background in the theory of infinite series with complex terms, it can be shown that the series in (D) converges for all real θ. (a) Recalling that i2 = −1, write enough terms of the sequence {in} to convince yourself that the sequence is repetitive: 1, i, −1, −i, 1, i, −1, −i, 1, i, −1, −i, 1, i, −1, −i, · · · . Use this to group the terms in (D) as eiθ = 1 −θ2 2 + θ4 4 + · · · + i θ −θ3 3! + θ5 5! + · · · = ∞ X n=0 (−1)n θ2n (2n)! + i ∞ X n=0 (−1)n θ2n+1 (2n + 1)!. By comparing this result with (B) and (C), conclude that eiθ = cos θ + i sin θ. (E) This is Euler’s identity. (b) Starting from eiθ1eiθ2 = (cos θ1 + i sin θ1)(cos θ2 + i sin θ2), collect the real part (the terms not multiplied by i) and the imaginary part (the terms multi- plied by i) on the right, and use the trigonometric identities cos(θ1 + θ2) = cos θ1 cos θ2 −sin θ1 sin θ2 sin(θ1 + θ2) = sin θ1 cos θ2 + cos θ1 sin θ2 to verify that ei(θ1+θ2) = eiθ1eiθ2, as you would expect from the use of the exponential notation eiθ. | Elementary Differential Equations with Boundary Value Problems_Page_230_Chunk2635 |
Section 5.3 Nonhomogeneous Linear Equations 221 (c) If α and β are real numbers, define eα+iβ = eαeiβ = eα(cos β + i sin β). (F) Show that if z1 = α1 + iβ1 and z2 = α2 + iβ2 then ez1+z2 = ez1ez2. (d) Let a, b, and c be real numbers, with a ̸= 0. Let z = u + iv where u and v are real-valued functions of x. Then we say that z is a solution of ay′′ + by′ + cy = 0 (G) if u and v are both solutions of (G). Use Theorem 5.2.1(c) to verify that if the characteristic equation of (G) has complex conjugate roots λ ± iω then z1 = e(λ+iω)x and z2 = e(λ−iω)x are both solutions of (G). 5.3 NONHOMOGENEOUS LINEAR EQUATIONS We’ll now consider the nonhomogeneous linear second order equation y′′ + p(x)y′ + q(x)y = f(x), (5.3.1) where the forcing function f isn’t identically zero. The next theorem, an extension of Theorem 5.1.1, gives sufficient conditions for existence and uniqueness of solutions of initial value problems for (5.3.1). We omit the proof, which is beyond the scope of this book. Theorem 5.3.1 Suppose p, , q and f are continuous on an open interval (a, b), let x0 be any point in (a, b), and let k0 and k1 be arbitrary real numbers. Then the initial value problem y′′ + p(x)y′ + q(x)y = f(x), y(x0) = k0, y′(x0) = k1 has a unique solution on (a, b). To find the general solution of (5.3.1) on an interval (a, b) where p, q, and f are continuous, it’s necessary to find the general solution of the associated homogeneous equation y′′ + p(x)y′ + q(x)y = 0 (5.3.2) on (a, b). We call (5.3.2) the complementary equation for (5.3.1). The next theorem shows how to find the general solution of (5.3.1) if we know one solution yp of (5.3.1) and a fundamental set of solutions of (5.3.2). We call yp a particular solution of (5.3.1); it can be any solution that we can find, one way or another. Theorem 5.3.2 Suppose p, q, and f are continuous on (a, b). Let yp be a particular solution of y′′ + p(x)y′ + q(x)y = f(x) (5.3.3) on (a, b), and let {y1, y2} be a fundamental set of solutions of the complementary equation y′′ + p(x)y′ + q(x)y = 0 (5.3.4) on (a, b). Then y is a solution of (5.3.3) on (a, b) if and only if y = yp + c1y1 + c2y2, (5.3.5) where c1 and c2 are constants. | Elementary Differential Equations with Boundary Value Problems_Page_231_Chunk2636 |
222 Chapter 5 Linear Second Order Equations Proof We first show that y in (5.3.5) is a solution of (5.3.3) for any choice of the constants c1 and c2. Differentiating (5.3.5) twice yields y′ = y′ p + c1y′ 1 + c2y′ 2 and y′′ = y′′ p + c1y′′ 1 + c2y′′ 2 , so y′′ + p(x)y′ + q(x)y = (y′′ p + c1y′′ 1 + c2y′′ 2 ) + p(x)(y′ p + c1y′ 1 + c2y′ 2) +q(x)(yp + c1y1 + c2y2) = (y′′ p + p(x)y′ p + q(x)yp) + c1(y′′ 1 + p(x)y′ 1 + q(x)y1) +c2(y′′ 2 + p(x)y′ 2 + q(x)y2) = f + c1 · 0 + c2 · 0 = f, since yp satisfies (5.3.3) and y1 and y2 satisfy (5.3.4). Now we’ll show that every solution of (5.3.3) has the form (5.3.5) for some choice of the constants c1 and c2. Suppose y is a solution of (5.3.3). We’ll show that y −yp is a solution of (5.3.4), and therefore of the form y −yp = c1y1 + c2y2, which implies (5.3.5). To see this, we compute (y −yp)′′ + p(x)(y −yp)′ + q(x)(y −yp) = (y′′ −y′′ p) + p(x)(y′ −y′ p) +q(x)(y −yp) = (y′′ + p(x)y′ + q(x)y) −(y′′ p + p(x)y′ p + q(x)yp) = f(x) −f(x) = 0, since y and yp both satisfy (5.3.3). We say that (5.3.5) is the general solution of (5.3.3) on (a, b). If P0, P1, and F are continuous and P0 has no zeros on (a, b), then Theorem 5.3.2 implies that the general solution of P0(x)y′′ + P1(x)y′ + P2(x)y = F (x) (5.3.6) on (a, b) is y = yp + c1y1 + c2y2, where yp is a particular solution of (5.3.6) on (a, b) and {y1, y2} is a fundamental set of solutions of P0(x)y′′ + P1(x)y′ + P2(x)y = 0 on (a, b). To see this, we rewrite (5.3.6) as y′′ + P1(x) P0(x)y′ + P2(x) P0(x)y = F (x) P0(x) and apply Theorem 5.3.2 with p = P1/P0, q = P2/P0, and f = F/P0. To avoid awkward wording in examples and exercises, we won’t specify the interval (a, b) when we ask for the general solution of a specific linear second order equation, or for a fundamental set of solutions of a homogeneous linear second order equation. Let’s agree that this always means that we want the general solution (or a fundamental set of solutions, as the case may be) on every open interval on which p, q, and f are continuous if the equation is of the form (5.3.3), or on which P0, P1, P2, and F are continuous and P0 has no zeros, if the equation is of the form (5.3.6). We leave it to you to identify these intervals in specific examples and exercises. For completeness, we point out that if P0, P1, P2, and F are all continuous on an open interval (a, b), but P0 does have a zero in (a, b), then (5.3.6) may fail to have a general solution on (a, b) in the sense just defined. Exercises 42–44 illustrate this point for a homogeneous equation. In this section we to limit ourselves to applications of Theorem 5.3.2 where we can guess at the form of the particular solution. | Elementary Differential Equations with Boundary Value Problems_Page_232_Chunk2637 |
Section 5.3 Nonhomogeneous Linear Equations 223 Example 5.3.1 (a) Find the general solution of y′′ + y = 1. (5.3.7) (b) Solve the initial value problem y′′ + y = 1, y(0) = 2, y′(0) = 7. (5.3.8) SOLUTION(a) We can apply Theorem 5.3.2 with (a, b) = (−∞, ∞), since the functions p ≡0, q ≡1, and f ≡1 in (5.3.7) are continuous on (−∞, ∞). By inspection we see that yp ≡1 is a particular solution of (5.3.7). Since y1 = cos x and y2 = sin x form a fundamental set of solutions of the complementary equation y′′ + y = 0, the general solution of (5.3.7) is y = 1 + c1 cos x + c2 sin x. (5.3.9) SOLUTION(b) Imposing the initial condition y(0) = 2 in (5.3.9) yields 2 = 1 + c1, so c1 = 1. Differen- tiating (5.3.9) yields y′ = −c1 sin x + c2 cos x. Imposing the initial condition y′(0) = 7 here yields c2 = 7, so the solution of (5.3.8) is y = 1 + cos x + 7 sinx. Figure 5.3.1 is a graph of this function. Example 5.3.2 (a) Find the general solution of y′′ −2y′ + y = −3 −x + x2. (5.3.10) (b) Solve the initial value problem y′′ −2y′ + y = −3 −x + x2, y(0) = −2, y′(0) = 1. (5.3.11) SOLUTION(a) The characteristic polynomial of the complementary equation y′′ −2y′ + y = 0 is r2 −2r + 1 = (r −1)2, so y1 = ex and y2 = xex form a fundamental set of solutions of the complementary equation. To guess a form for a particular solution of (5.3.10), we note that substituting a second degree polynomial yp = A + Bx + Cx2 into the left side of (5.3.10) will produce another second degree polynomial with coefficients that depend upon A, B, and C. The trick is to choose A, B, and C so the polynomials on the two sides of (5.3.10) have the same coefficients; thus, if yp = A + Bx + Cx2 then y′ p = B + 2Cx and y′′ p = 2C, so y′′ p −2y′ p + yp = 2C −2(B + 2Cx) + (A + Bx + Cx2) = (2C −2B + A) + (−4C + B)x + Cx2 = −3 −x + x2. | Elementary Differential Equations with Boundary Value Problems_Page_233_Chunk2638 |
224 Chapter 5 Linear Second Order Equations 1 2 3 4 5 6 2 4 6 8 − 2 − 4 − 6 − 8 x y Figure 5.3.1 y = 1 + cos x + 7 sinx Equating coefficients of like powers of x on the two sides of the last equality yields C = 1 B −4C = −1 A −2B + 2C = −3, so C = 1, B = −1 + 4C = 3, and A = −3 −2C + 2B = 1. Therefore yp = 1 + 3x + x2 is a particular solution of (5.3.10) and Theorem 5.3.2 implies that y = 1 + 3x + x2 + ex(c1 + c2x) (5.3.12) is the general solution of (5.3.10). SOLUTION(b) Imposing the initial condition y(0) = −2 in (5.3.12) yields −2 = 1 + c1, so c1 = −3. Differentiating (5.3.12) yields y′ = 3 + 2x + ex(c1 + c2x) + c2ex, and imposing the initial condition y′(0) = 1 here yields 1 = 3 + c1 + c2, so c2 = 1. Therefore the solution of (5.3.11) is y = 1 + 3x + x2 −ex(3 −x). Figure 5.3.2 is a graph of this solution. Example 5.3.3 Find the general solution of x2y′′ + xy′ −4y = 2x4 (5.3.13) on (−∞, 0) and (0, ∞). | Elementary Differential Equations with Boundary Value Problems_Page_234_Chunk2639 |
Section 5.3 Nonhomogeneous Linear Equations 225 0.5 1.0 1.5 2.0 2 4 6 8 10 12 14 16 x y Figure 5.3.2 y = 1 + 3x + x2 −ex(3 −x) Solution In Example 5.1.3, we verified that y1 = x2 and y2 = 1/x2 form a fundamental set of solutions of the complementary equation x2y′′ + xy′ −4y = 0 on (−∞, 0) and (0, ∞). To find a particular solution of (5.3.13), we note that if yp = Ax4, where A is a constant then both sides of (5.3.13) will be constant multiples of x4 and we may be able to choose A so the two sides are equal. This is true in this example, since if yp = Ax4 then x2y′′ p + xy′ p −4yp = x2(12Ax2) + x(4Ax3) −4Ax4 = 12Ax4 = 2x4 if A = 1/6; therefore, yp = x4/6 is a particular solution of (5.3.13) on (−∞, ∞). Theorem 5.3.2 implies that the general solution of (5.3.13) on (−∞, 0) and (0, ∞) is y = x4 6 + c1x2 + c2 x2 . The Principle of Superposition The next theorem enables us to break a nonhomogeous equation into simpler parts, find a particular solution for each part, and then combine their solutions to obtain a particular solution of the original problem. Theorem 5.3.3 [The Principle of Superposition] Suppose yp1 is a particular solution of y′′ + p(x)y′ + q(x)y = f1(x) on (a, b) and yp2 is a particular solution of y′′ + p(x)y′ + q(x)y = f2(x) | Elementary Differential Equations with Boundary Value Problems_Page_235_Chunk2640 |
226 Chapter 5 Linear Second Order Equations on (a, b). Then yp = yp1 + yp2 is a particular solution of y′′ + p(x)y′ + q(x)y = f1(x) + f2(x) on (a, b). Proof If yp = yp1 + yp2 then y′′ p + p(x)y′ p + q(x)yp = (yp1 + yp2)′′ + p(x)(yp1 + yp2)′ + q(x)(yp1 + yp2) = | Elementary Differential Equations with Boundary Value Problems_Page_236_Chunk2641 |
Section 5.3 Nonhomogeneous Linear Equations 227 Solution The right side F (x) = 2x4 + 4x2 in (5.3.17) is the sum of the right sides F1(x) = 2x4 and F2(x) = 4x2. in (5.3.15) and (5.3.16). Therefore the principle of superposition implies that yp = yp1 + yp2 = x4 15 + x2 3 is a particular solution of (5.3.17). 5.3 Exercises In Exercises 1–6 find a particular solution by the method used in Example 5.3.2. Then find the general solution and, where indicated, solve the initial value problem and graph the solution. 1. y′′ + 5y′ −6y = 22 + 18x −18x2 2. y′′ −4y′ + 5y = 1 + 5x 3. y′′ + 8y′ + 7y = −8 −x + 24x2 + 7x3 4. y′′ −4y′ + 4y = 2 + 8x −4x2 5. C/G y′′ + 2y′ + 10y = 4 + 26x + 6x2 + 10x3, y(0) = 2, y′(0) = 9 6. C/G y′′ + 6y′ + 10y = 22 + 20x, y(0) = 2, y′(0) = −2 7. Show that the method used in Example 5.3.2 won’t yield a particular solution of y′′ + y′ = 1 + 2x + x2; (A) that is, (A) does’nt have a particular solution of the form yp = A + Bx + Cx2, where A, B, and C are constants. In Exercises 8–13 find a particular solution by the method used in Example 5.3.3. 8. x2y′′ + 7xy′ + 8y = 6 x 9. x2y′′ −7xy′ + 7y = 13x1/2 10. x2y′′ −xy′ + y = 2x3 11. x2y′′ + 5xy′ + 4y = 1 x3 12. x2y′′ + xy′ + y = 10x1/3 13. x2y′′ −3xy′ + 13y = 2x4 14. Show that the method suggested for finding a particular solution in Exercises 8-13 won’t yield a particular solution of x2y′′ + 3xy′ −3y = 1 x3 ; (A) that is, (A) doesn’t have a particular solution of the form yp = A/x3. 15. Prove: If a, b, c, α, and M are constants and M ̸= 0 then ax2y′′ + bxy′ + cy = Mxα has a particular solution yp = Axα (A = constant) if and only if aα(α −1) + bα + c ̸= 0. | Elementary Differential Equations with Boundary Value Problems_Page_237_Chunk2642 |
228 Chapter 5 Linear Second Order Equations If a, b, c, and α are constants, then a(eαx)′′ + b(eαx)′ + ceαx = (aα2 + bα + c)eαx. Use this in Exercises 16–21 to find a particular solution . Then find the general solution and, where indicated, solve the initial value problem and graph the solution. 16. y′′ + 5y′ −6y = 6e3x 17. y′′ −4y′ + 5y = e2x 18. C/G y′′ + 8y′ + 7y = 10e−2x, y(0) = −2, y′(0) = 10 19. C/G y′′ −4y′ + 4y = ex, y(0) = 2, y′(0) = 0 20. y′′ + 2y′ + 10y = ex/2 21. y′′ + 6y′ + 10y = e−3x 22. Show that the method suggested for finding a particular solution in Exercises 16-21 won’t yield a particular solution of y′′ −7y′ + 12y = 5e4x; (A) that is, (A) doesn’t have a particular solution of the form yp = Ae4x. 23. Prove: If α and M are constants and M ̸= 0 then constant coefficient equation ay′′ + by′ + cy = Meαx has a particular solution yp = Aeαx (A = constant) if and only if eαx isn’t a solution of the complementary equation. If ω is a constant, differentiating a linear combination of cos ωx and sin ωx with respect to x yields another linear combination of cos ωx and sin ωx. In Exercises 24–29 use this to find a particular solution of the equation. Then find the general solution and, where indicated, solve the initial value problem and graph the solution. 24. y′′ −8y′ + 16y = 23 cos x −7 sin x 25. y′′ + y′ = −8 cos 2x + 6 sin2x 26. y′′ −2y′ + 3y = −6 cos 3x + 6 sin 3x 27. y′′ + 6y′ + 13y = 18 cos x + 6 sin x 28. C/G y′′ + 7y′ + 12y = −2 cos 2x + 36 sin 2x, y(0) = −3, y′(0) = 3 29. C/G y′′ −6y′ + 9y = 18 cos 3x + 18 sin3x, y(0) = 2, y′(0) = 2 30. Find the general solution of y′′ + ω2 0y = M cos ωx + N sin ωx, where M and N are constants and ω and ω0 are distinct positive numbers. 31. Show that the method suggested for finding a particular solution in Exercises 24-29 won’t yield a particular solution of y′′ + y = cos x + sin x; (A) that is, (A) does not have a particular solution of the form yp = A cos x + B sin x. | Elementary Differential Equations with Boundary Value Problems_Page_238_Chunk2643 |
Section 5.4 The Method of Undetermined Coefficients 229 32. Prove: If M, N are constants (not both zero) and ω > 0, the constant coefficient equation ay′′ + by′ + cy = M cos ωx + N sin ωx (A) has a particular solution that’s a linear combination of cos ωx and sin ωx if and only if the left side of (A) is not of the form a(y′′+ω2y), so that cos ωx and sin ωx are solutionsof the complementary equation. In Exercises 33–38 refer to the cited exercises and use the principal of superposition to find a particular solution. Then find the general solution. 33. y′′ + 5y′ −6y = 22 + 18x −18x2 + 6e3x (See Exercises 1 and 16.) 34. y′′ −4y′ + 5y = 1 + 5x + e2x (See Exercises 2 and 17.) 35. y′′ + 8y′ + 7y = −8 −x + 24x2 + 7x3 + 10e−2x (See Exercises 3 and 18.) 36. y′′ −4y′ + 4y = 2 + 8x −4x2 + ex (See Exercises 4 and 19.) 37. y′′ + 2y′ + 10y = 4 + 26x + 6x2 + 10x3 + ex/2 (See Exercises 5 and 20.) 38. y′′ + 6y′ + 10y = 22 + 20x + e−3x (See Exercises 6 and 21.) 39. Prove: If yp1 is a particular solution of P0(x)y′′ + P1(x)y′ + P2(x)y = F1(x) on (a, b) and yp2 is a particular solution of P0(x)y′′ + P1(x)y′ + P2(x)y = F2(x) on (a, b), then yp = yp1 + yp2 is a solution of P0(x)y′′ + P1(x)y′ + P2(x)y = F1(x) + F2(x) on (a, b). 40. Suppose p, q, and f are continuous on (a, b). Let y1, y2, and yp be twice differentiable on (a, b), such that y = c1y1 + c2y2 + yp is a solution of y′′ + p(x)y′ + q(x)y = f on (a, b) for every choice of the constants c1, c2. Show that y1 and y2 are solutions of the comple- mentary equation on (a, b). 5.4 THE METHOD OF UNDETERMINED COEFFICIENTS I In this section we consider the constant coefficient equation ay′′ + by′ + cy = eαxG(x), (5.4.1) where α is a constant and G is a polynomial. From Theorem 5.3.2, the general solution of (5.4.1) is y = yp + c1y1 + c2y2, where yp is a particular solution of (5.4.1) and {y1, y2} is a fundamental set of solutions of the complementary equation ay′′ + by′ + cy = 0. In Section 5.2 we showed how to find {y1, y2}. In this section we’ll show how to find yp. The procedure that we’ll use is called the method of undetermined coefficients. Our first example is similar to Exercises 16–21. | Elementary Differential Equations with Boundary Value Problems_Page_239_Chunk2644 |
230 Chapter 5 Linear Second Order Equations I Example 5.4.1 Find a particular solution of y′′ −7y′ + 12y = 4e2x. (5.4.2) Then find the general solution. Solution Substituting yp = Ae2x for y in (5.4.2) will produce a constant multiple of Ae2x on the left side of (5.4.2), so it may be possible to choose A so that yp is a solution of (5.4.2). Let’s try it; if yp = Ae2x then y′′ p −7y′ p + 12yp = 4Ae2x −14Ae2x + 12Ae2x = 2Ae2x = 4e2x if A = 2. Therefore yp = 2e2x is a particular solution of (5.4.2). To find the general solution, we note that the characteristic polynomial of the complementary equation y′′ −7y′ + 12y = 0 (5.4.3) is p(r) = r2 −7r + 12 = (r −3)(r −4), so {e3x, e4x} is a fundamental set of solutions of (5.4.3). Therefore the general solution of (5.4.2) is y = 2e2x + c1e3x + c2e4x. Example 5.4.2 Find a particular solution of y′′ −7y′ + 12y = 5e4x. (5.4.4) Then find the general solution. Solution Fresh from our success in finding a particular solution of (5.4.2) — where we chose yp = Ae2x because the right side of (5.4.2) is a constant multiple of e2x — it may seem reasonable to try yp = Ae4x as a particular solution of (5.4.4). However, this won’t work, since we saw in Example 5.4.1 that e4x is a solution of the complementary equation (5.4.3), so substituting yp = Ae4x into the left side of (5.4.4) produces zero on the left, no matter how we chooseA. To discover a suitable form for yp, we use the same approach that we used in Section 5.2 to find a second solution of ay′′ + by′ + cy = 0 in the case where the characteristic equation has a repeated real root: we look for solutions of (5.4.4) in the form y = ue4x, where u is a function to be determined. Substituting y = ue4x, y′ = u′e4x + 4ue4x, and y′′ = u′′e4x + 8u′e4x + 16ue4x (5.4.5) into (5.4.4) and canceling the common factor e4x yields (u′′ + 8u′ + 16u) −7(u′ + 4u) + 12u = 5, or u′′ + u′ = 5. By inspection we see that up = 5x is a particular solution of this equation, so yp = 5xe4x is a particular solution of (5.4.4). Therefore y = 5xe4x + c1e3x + c2e4x is the general solution. | Elementary Differential Equations with Boundary Value Problems_Page_240_Chunk2645 |
Section 5.4 The Method of Undetermined Coefficients 231 Example 5.4.3 Find a particular solution of y′′ −8y′ + 16y = 2e4x. (5.4.6) Solution Since the characteristic polynomial of the complementary equation y′′ −8y′ + 16y = 0 (5.4.7) is p(r) = r2 −8r + 16 = (r −4)2, both y1 = e4x and y2 = xe4x are solutions of (5.4.7). Therefore (5.4.6) does not have a solution of the form yp = Ae4x or yp = Axe4x. As in Example 5.4.2, we look for solutions of (5.4.6) in the form y = ue4x, where u is a function to be determined. Substituting from (5.4.5) into (5.4.6) and canceling the common factor e4x yields (u′′ + 8u′ + 16u) −8(u′ + 4u) + 16u = 2, or u′′ = 2. Integrating twice and taking the constants of integration to be zero shows that up = x2 is a particular solution of this equation, so yp = x2e4x is a particular solution of (5.4.4). Therefore y = e4x(x2 + c1 + c2x) is the general solution. The preceding examples illustrate the following facts concerning the form of a particular solution yp of a constant coefficent equation ay′′ + by′ + cy = keαx, where k is a nonzero constant: (a) If eαx isn’t a solution of the complementary equation ay′′ + by′ + cy = 0, (5.4.8) then yp = Aeαx, where A is a constant. (See Example 5.4.1). (b) If eαx is a solution of (5.4.8) but xeαx is not, then yp = Axeαx, where A is a constant. (See Example 5.4.2.) (c) If both eαx and xeαx are solutions of (5.4.8), then yp = Ax2eαx, where A is a constant. (See Example 5.4.3.) See Exercise 30 for the proofs of these facts. In all three cases you can just substitute the appropriate form for yp and its derivatives directly into ay′′ p + by′ p + cyp = keαx, and solve for the constant A, as we did in Example 5.4.1. (See Exercises 31–33.) However, if the equation is ay′′ + by′ + cy = keαxG(x), where G is a polynomial of degree greater than zero, we recommend that you use the substitution y = ueαx as we did in Examples 5.4.2 and 5.4.3. The equation for u will turn out to be au′′ + p′(α)u′ + p(α)u = G(x), (5.4.9) | Elementary Differential Equations with Boundary Value Problems_Page_241_Chunk2646 |
232 Chapter 5 Linear Second Order Equations I where p(r) = ar2 + br + c is the characteristic polynomial of the complementary equation and p′(r) = 2ar + b (Exercise 30); however, you shouldn’t memorize this since it’s easy to derive the equation for u in any particular case. Note, however, that if eαx is a solution of the complementary equation then p(α) = 0, so (5.4.9) reduces to au′′ + p′(α)u′ = G(x), while if both eαx and xeαx are solutions of the complementary equation then p(r) = a(r −α)2 and p′(r) = 2a(r −α), so p(α) = p′(α) = 0 and (5.4.9) reduces to au′′ = G(x). Example 5.4.4 Find a particular solution of y′′ −3y′ + 2y = e3x(−1 + 2x + x2). (5.4.10) Solution Substituting y = ue3x, y′ = u′e3x + 3ue3x, and y′′ = u′′e3x + 6u′e3x + 9ue3x into (5.4.10) and canceling e3x yields (u′′ + 6u′ + 9u) −3(u′ + 3u) + 2u = −1 + 2x + x2, or u′′ + 3u′ + 2u = −1 + 2x + x2. (5.4.11) As in Example 2, in order to guess a form for a particular solution of (5.4.11), we note that substituting a second degree polynomial up = A + Bx + Cx2 for u in the left side of (5.4.11) produces another second degree polynomial with coefficients that depend upon A, B, and C; thus, if up = A + Bx + Cx2 then u′ p = B + 2Cx and u′′ p = 2C. If up is to satisfy (5.4.11), we must have u′′ p + 3u′ p + 2up = 2C + 3(B + 2Cx) + 2(A + Bx + Cx2) = (2C + 3B + 2A) + (6C + 2B)x + 2Cx2 = −1 + 2x + x2. Equating coefficients of like powers of x on the two sides of the last equality yields 2C = 1 2B + 6C = 2 2A + 3B + 2C = −1. Solving these equations for C, B, and A (in that order) yields C = 1/2, B = −1/2, A = −1/4. Therefore up = −1 4(1 + 2x −2x2) is a particular solution of (5.4.11), and yp = upe3x = −e3x 4 (1 + 2x −2x2) is a particular solution of (5.4.10). | Elementary Differential Equations with Boundary Value Problems_Page_242_Chunk2647 |
Section 5.4 The Method of Undetermined Coefficients 233 Example 5.4.5 Find a particular solution of y′′ −4y′ + 3y = e3x(6 + 8x + 12x2). (5.4.12) Solution Substituting y = ue3x, y′ = u′e3x + 3ue3x, and y′′ = u′′e3x + 6u′e3x + 9ue3x into (5.4.12) and canceling e3x yields (u′′ + 6u′ + 9u) −4(u′ + 3u) + 3u = 6 + 8x + 12x2, or u′′ + 2u′ = 6 + 8x + 12x2. (5.4.13) There’s no u term in this equation, since e3x is a solution of the complementary equation for (5.4.12). (See Exercise 30.) Therefore (5.4.13) does not have a particular solution of the form up = A+Bx+Cx2 that we used successfully in Example 5.4.4, since with this choice of up, u′′ p + 2u′ p = 2C + (B + 2Cx) can’t contain the last term (12x2) on the right side of (5.4.13). Instead, let’s try up = Ax + Bx2 + Cx3 on the grounds that u′ p = A + 2Bx + 3Cx2 and u′′ p = 2B + 6Cx together contain all the powers of x that appear on the right side of (5.4.13). Substituting these expressions in place of u′ and u′′ in (5.4.13) yields (2B + 6Cx) + 2(A + 2Bx + 3Cx2) = (2B + 2A) + (6C + 4B)x + 6Cx2 = 6 + 8x + 12x2. Comparing coefficients of like powers of x on the two sides of the last equality shows that up satisfies (5.4.13) if 6C = 12 4B + 6C = 8 2A + 2B = 6. Solving these equations successively yields C = 2, B = −1, and A = 4. Therefore up = x(4 −x + 2x2) is a particular solution of (5.4.13), and yp = upe3x = xe3x(4 −x + 2x2) is a particular solution of (5.4.12). Example 5.4.6 Find a particular solution of 4y′′ + 4y′ + y = e−x/2(−8 + 48x + 144x2). (5.4.14) Solution Substituting y = ue−x/2, y′ = u′e−x/2 −1 2ue−x/2, and y′′ = u′′e−x/2 −u′e−x/2 + 1 4ue−x/2 | Elementary Differential Equations with Boundary Value Problems_Page_243_Chunk2648 |
234 Chapter 5 Linear Second Order Equations I into (5.4.14) and canceling e−x/2 yields 4 u′′ −u′ + u 4 + 4 u′ −u 2 + u = 4u′′ = −8 + 48x + 144x2, or u′′ = −2 + 12x + 36x2, (5.4.15) which does not contain u or u′ because e−x/2 and xe−x/2 are both solutions of the complementary equation. (See Exercise 30.) To obtain a particular solution of (5.4.15) we integrate twice, taking the constants of integration to be zero; thus, u′ p = −2x + 6x2 + 12x3 and up = −x2 + 2x3 + 3x4 = x2(−1 + 2x + 3x2). Therefore yp = upe−x/2 = x2e−x/2(−1 + 2x + 3x2) is a particular solution of (5.4.14). Summary The preceding examples illustrate the following facts concerning particular solutions of a constant coef- ficent equation of the form ay′′ + by′ + cy = eαxG(x), where G is a polynomial (see Exercise 30): (a) If eαx isn’t a solution of the complementary equation ay′′ + by′ + cy = 0, (5.4.16) then yp = eαxQ(x), where Q is a polynomial of the same degree as G. (See Example 5.4.4). (b) If eαx is a solution of (5.4.16) but xeαx is not, then yp = xeαxQ(x), where Q is a polynomial of the same degree as G. (See Example 5.4.5.) (c) If both eαx and xeαx are solutions of (5.4.16), then yp = x2eαxQ(x), where Q is a polynomial of the same degree as G. (See Example 5.4.6.) In all three cases, you can just substitute the appropriate form for yp and its derivatives directly into ay′′ p + by′ p + cyp = eαxG(x), and solve for the coefficients of the polynomial Q. However, if you try this you will see that the compu- tations are more tedious than those that you encounter by making the substitution y = ueαx and finding a particular solution of the resulting equation for u. (See Exercises 34-36.) In Case (a) the equation for u will be of the form au′′ + p′(α)u′ + p(α)u = G(x), with a particular solution of the form up = Q(x), a polynomial of the same degree as G, whose coeffi- cients can be found by the method used in Example 5.4.4. In Case (b) the equation for u will be of the form au′′ + p′(α)u′ = G(x) (no u term on the left), with a particular solution of the form up = xQ(x), where Q is a polynomial of the same degree as G whose coefficents can be found by the method used in Example 5.4.5. In Case (c) the equation for u will be of the form au′′ = G(x) | Elementary Differential Equations with Boundary Value Problems_Page_244_Chunk2649 |
Section 5.4 The Method of Undetermined Coefficients 235 with a particular solution of the form up = x2Q(x) that can be obtained by integrating G(x)/a twice and taking the constants of integration to be zero, as in Example 5.4.6. Using the Principle of Superposition The next example shows how to combine the method of undetermined coefficients and Theorem 5.3.3, the principle of superposition. Example 5.4.7 Find a particular solution of y′′ −7y′ + 12y = 4e2x + 5e4x. (5.4.17) Solution In Example 5.4.1 we found that yp1 = 2e2x is a particular solution of y′′ −7y′ + 12y = 4e2x, and in Example 5.4.2 we found that yp2 = 5xe4x is a particular solution of y′′ −7y′ + 12y = 5e4x. Therefore the principle of superposition implies that yp = 2e2x+5xe4x is a particular solution of (5.4.17). 5.4 Exercises In Exercises 1–14 find a particular solution. 1. y′′ −3y′ + 2y = e3x(1 + x) 2. y′′ −6y′ + 5y = e−3x(35 −8x) 3. y′′ −2y′ −3y = ex(−8 + 3x) 4. y′′ + 2y′ + y = e2x(−7 −15x + 9x2) 5. y′′ + 4y = e−x(7 −4x + 5x2) 6. y′′ −y′ −2y = ex(9 + 2x −4x2) 7. y′′ −4y′ −5y = −6xe−x 8. y′′ −3y′ + 2y = ex(3 −4x) 9. y′′ + y′ −12y = e3x(−6 + 7x) 10. 2y′′ −3y′ −2y = e2x(−6 + 10x) 11. y′′ + 2y′ + y = e−x(2 + 3x) 12. y′′ −2y′ + y = ex(1 −6x) 13. y′′ −4y′ + 4y = e2x(1 −3x + 6x2) 14. 9y′′ + 6y′ + y = e−x/3(2 −4x + 4x2) In Exercises 15–19 find the general solution. 15. y′′ −3y′ + 2y = e3x(1 + x) 16. y′′ −6y′ + 8y = ex(11 −6x) 17. y′′ + 6y′ + 9y = e2x(3 −5x) 18. y′′ + 2y′ −3y = −16xex 19. y′′ −2y′ + y = ex(2 −12x) | Elementary Differential Equations with Boundary Value Problems_Page_245_Chunk2650 |
236 Chapter 5 Linear Second Order Equations I In Exercises 20–23 solve the initial value problem and plot the solution. 20. C/G y′′ −4y′ −5y = 9e2x(1 + x), y(0) = 0, y′(0) = −10 21. C/G y′′ + 3y′ −4y = e2x(7 + 6x), y(0) = 2, y′(0) = 8 22. C/G y′′ + 4y′ + 3y = −e−x(2 + 8x), y(0) = 1, y′(0) = 2 23. C/G y′′ −3y′ −10y = 7e−2x, y(0) = 1, y′(0) = −17 In Exercises 24–29 use the principle of superposition to find a particular solution. 24. y′′ + y′ + y = xex + e−x(1 + 2x) 25. y′′ −7y′ + 12y = −ex(17 −42x) −e3x 26. y′′ −8y′ + 16y = 6xe4x + 2 + 16x + 16x2 27. y′′ −3y′ + 2y = −e2x(3 + 4x) −ex 28. y′′ −2y′ + 2y = ex(1 + x) + e−x(2 −8x + 5x2) 29. y′′ + y = e−x(2 −4x + 2x2) + e3x(8 −12x −10x2) 30. (a) Prove that y is a solution of the constant coefficient equation ay′′ + by′ + cy = eαxG(x) (A) if and only if y = ueαx, where u satisfies au′′ + p′(α)u′ + p(α)u = G(x) (B) and p(r) = ar2 + br + c is the characteristic polynomial of the complementary equation ay′′ + by′ + cy = 0. For the rest of this exercise, let G be a polynomial. Give the requested proofs for the case where G(x) = g0 + g1x + g2x2 + g3x3. (b) Prove that if eαx isn’t a solution of the complementary equation then (B) has a particular solution of the form up = A(x), where A is a polynomial of the same degree as G, as in Example 5.4.4. Conclude that (A) has a particular solution of the form yp = eαxA(x). (c) Show that if eαx is a solution of the complementary equation and xeαx isn’t , then (B) has a particular solution of the form up = xA(x), where A is a polynomial of the same degree as G, as in Example 5.4.5. Conclude that (A) has a particular solution of the form yp = xeαxA(x). (d) Show that if eαx and xeαx are both solutions of the complementary equation then (B) has a particular solution of the form up = x2A(x), where A is a polynomial of the same degree as G, and x2A(x) can be obtained by integrating G/a twice, taking the constants of integration to be zero, as in Example 5.4.6. Conclude that (A) has a particular solution of the form yp = x2eαxA(x). Exercises 31–36 treat the equations considered in Examples 5.4.1–5.4.6. Substitute the suggested form of yp into the equation and equate the resulting coefficients of like functions on the two sides of the resulting equation to derive a set of simultaneous equations for the coefficients in yp. Then solve for the coefficients to obtain yp. Compare the work you’ve done with the work required to obtain the same results in Examples 5.4.1–5.4.6. | Elementary Differential Equations with Boundary Value Problems_Page_246_Chunk2651 |
Section 5.4 The Method of Undetermined Coefficients 237 31. Compare with Example 5.4.1: y′′ −7y′ + 12y = 4e2x; yp = Ae2x 32. Compare with Example 5.4.2: y′′ −7y′ + 12y = 5e4x; yp = Axe4x 33. Compare with Example 5.4.3. y′′ −8y′ + 16y = 2e4x; yp = Ax2e4x 34. Compare with Example 5.4.4: y′′ −3y′ + 2y = e3x(−1 + 2x + x2), yp = e3x(A + Bx + Cx2) 35. Compare with Example 5.4.5: y′′ −4y′ + 3y = e3x(6 + 8x + 12x2), yp = e3x(Ax + Bx2 + Cx3) 36. Compare with Example 5.4.6: 4y′′ + 4y′ + y = e−x/2(−8 + 48x + 144x2), yp = e−x/2(Ax2 + Bx3 + Cx4) 37. Write y = ueαx to find the general solution. (a) y′′ + 2y′ + y = e−x √x (b) y′′ + 6y′ + 9y = e−3x ln x (c) y′′ −4y′ + 4y = e2x 1 + x (d) 4y′′ + 4y′ + y = 4e−x/2 1 x + x 38. Suppose α ̸= 0 and k is a positive integer. In most calculus books integrals like R xkeαx dx are evaluated by integrating by parts k times. This exercise presents another method. Let y = Z eαxP (x) dx with P (x) = p0 + p1x + · · · + pkxk, (where pk ̸= 0). (a) Show that y = eαxu, where u′ + αu = P (x). (A) (b) Show that (A) has a particular solution of the form up = A0 + A1x + · · · + Akxk, where Ak, Ak−1, ..., A0 can be computed successively by equating coefficients of xk, xk−1, . . ., 1 on both sides of the equation u′ p + αup = P (x). (c) Conclude that Z eαxP (x) dx = | Elementary Differential Equations with Boundary Value Problems_Page_247_Chunk2652 |
238 Chapter 5 Linear Second Order Equations 39. Use the method of Exercise 38 to evaluate the integral. (a) R ex(4 + x) dx (b) R e−x(−1 + x2) dx (c) R x3e−2x dx (d) R ex(1 + x)2 dx (e) R e3x(−14 + 30x + 27x2) dx (f) R e−x(1 + 6x2 −14x3 + 3x4) dx 40. Use the method suggested in Exercise 38 to evaluate R xkeαx dx, where k is an arbitrary positive integer and α ̸= 0. 5.5 THE METHOD OF UNDETERMINED COEFFICIENTS II In this section we consider the constant coefficient equation ay′′ + by′ + cy = eλx (P (x) cos ωx + Q(x) sinωx) (5.5.1) where λ and ω are real numbers, ω ̸= 0, and P and Q are polynomials. We want to find a particular solution of (5.5.1). As in Section 5.4, the procedure that we will use is called the method of undetermined coefficients. Forcing Functions Without Exponential Factors We begin with the case where λ = 0 in (5.5.1); thus, we we want to find a particular solution of ay′′ + by′ + cy = P (x) cos ωx + Q(x) sinωx, (5.5.2) where P and Q are polynomials. Differentiating xr cos ωx and xr sin ωx yields d dxxr cos ωx = −ωxr sin ωx + rxr−1 cos ωx and d dxxr sin ωx = ωxr cos ωx + rxr−1 sin ωx. This implies that if yp = A(x) cos ωx + B(x) sin ωx where A and B are polynomials, then ay′′ p + by′ p + cyp = F (x) cos ωx + G(x) sinωx, where F and G are polynomials with coefficients that can be expressed in terms of the coefficients of A and B. This suggests that we try to choose A and B so that F = P and G = Q, respectively. Then yp will be a particular solution of (5.5.2). The next theorem tells us how to choose the proper form for yp. For the proof see Exercise 37. Theorem 5.5.1 Suppose ω is a positive number and P and Q are polynomials. Let k be the larger of the degrees of P and Q. Then the equation ay′′ + by′ + cy = P (x) cos ωx + Q(x) sin ωx has a particular solution yp = A(x) cos ωx + B(x) sin ωx, (5.5.3) | Elementary Differential Equations with Boundary Value Problems_Page_248_Chunk2653 |
Section 5.5 The Method of Undetermined Coefficients II 239 where A(x) = A0 + A1x + · · · + Akxk and B(x) = B0 + B1x + · · · + Bkxk, provided that cos ωx and sinωx are not solutions of the complementary equation. The solutions of a(y′′ + ω2y) = P (x) cos ωx + Q(x) sinωx (for which cos ωx and sin ωx are solutions of the complementary equation) are of the form (5.5.3), where A(x) = A0x + A1x2 + · · · + Akxk+1 and B(x) = B0x + B1x2 + · · · + Bkxk+1. For an analog of this theorem that’s applicable to (5.5.1), see Exercise 38. Example 5.5.1 Find a particular solution of y′′ −2y′ + y = 5 cos 2x + 10 sin2x. (5.5.4) Solution In (5.5.4) the coefficients of cos 2x and sin2x are both zero degree polynomials (constants). Therefore Theorem 5.5.1 implies that (5.5.4) has a particular solution yp = A cos 2x + B sin 2x. Since y′ p = −2A sin 2x + 2B cos 2x and y′′ p = −4(A cos 2x + B sin 2x), replacing y by yp in (5.5.4) yields y′′ p −2y′ p + yp = −4(A cos 2x + B sin 2x) −4(−A sin 2x + B cos 2x) +(A cos 2x + B sin 2x) = (−3A −4B) cos 2x + (4A −3B) sin 2x. Equating the coefficients of cos 2x and sin 2x here with the corresponding coefficients on the right side of (5.5.4) shows that yp is a solution of (5.5.4) if −3A −4B = 5 4A −3B = 10. Solving these equations yields A = 1, B = −2. Therefore yp = cos 2x −2 sin2x is a particular solution of (5.5.4). Example 5.5.2 Find a particular solution of y′′ + 4y = 8 cos 2x + 12 sin2x. (5.5.5) Solution The procedure used in Example 5.5.1 doesn’t work here; substituting yp = A cos 2x+B sin 2x for y in (5.5.5) yields y′′ p + 4yp = −4(A cos 2x + B sin 2x) + 4(A cos 2x + B sin 2x) = 0 | Elementary Differential Equations with Boundary Value Problems_Page_249_Chunk2654 |
240 Chapter 5 Linear Second Order Equations for any choice of A and B, since cos 2x and sin 2x are both solutions of the complementary equation for (5.5.5). We’re dealing with the second case mentioned in Theorem 5.5.1, and should therefore try a particular solution of the form yp = x(A cos 2x + B sin 2x). (5.5.6) Then y′ p = A cos 2x + B sin 2x + 2x(−A sin 2x + B cos 2x) and y′′ p = −4A sin 2x + 4B cos 2x −4x(A cos 2x + B sin 2x) = −4A sin 2x + 4B cos 2x −4yp (see (5.5.6)), so y′′ p + 4yp = −4A sin 2x + 4B cos 2x. Therefore yp is a solution of (5.5.5) if −4A sin 2x + 4B cos 2x = 8 cos 2x + 12 sin2x, which holds if A = −3 and B = 2. Therefore yp = −x(3 cos 2x −2 sin2x) is a particular solution of (5.5.5). Example 5.5.3 Find a particular solution of y′′ + 3y′ + 2y = (16 + 20x) cosx + 10 sinx. (5.5.7) Solution The coefficients of cos x and sin x in (5.5.7) are polynomials of degree one and zero, respec- tively. Therefore Theorem 5.5.1 tells us to look for a particular solution of (5.5.7) of the form yp = (A0 + A1x) cos x + (B0 + B1x) sinx. (5.5.8) Then y′ p = (A1 + B0 + B1x) cos x + (B1 −A0 −A1x) sin x (5.5.9) and y′′ p = (2B1 −A0 −A1x) cos x −(2A1 + B0 + B1x) sinx, (5.5.10) so y′′ p + 3y′ p + 2yp = [A0 + 3A1 + 3B0 + 2B1 + (A1 + 3B1)x] cos x + [B0 + 3B1 −3A0 −2A1 + (B1 −3A1)x] sin x. (5.5.11) Comparing the coefficients of x cos x, x sin x, cos x, and sin x here with the corresponding coefficients in (5.5.7) shows that yp is a solution of (5.5.7) if A1 + 3B1 = 20 −3A1 + B1 = 0 A0 + 3B0 + 3A1 + 2B1 = 16 −3A0 + B0 −2A1 + 3B1 = 10. Solving the first two equations yields A1 = 2, B1 = 6. Substituting these into the last two equations yields A0 + 3B0 = 16 −3A1 −2B1 = −2 −3A0 + B0 = 10 + 2A1 −3B1 = −4. | Elementary Differential Equations with Boundary Value Problems_Page_250_Chunk2655 |
Section 5.5 The Method of Undetermined Coefficients II 241 Solving these equations yields A0 = 1, B0 = −1. Substituting A0 = 1, A1 = 2, B0 = −1, B1 = 6 into (5.5.8) shows that yp = (1 + 2x) cos x −(1 −6x) sin x is a particular solution of (5.5.7). A Useful Observation In (5.5.9), (5.5.10), and (5.5.11) the polynomialsmultiplyingsin x can be obtained by replacing A0, A1, B0, and B1 by B0, B1, −A0, and −A1, respectively, in the polynomials mutiplying cos x. An analogous re- sult applies in general, as follows (Exercise 36). Theorem 5.5.2 If yp = A(x) cos ωx + B(x) sin ωx, where A(x) and B(x) are polynomials with coefficients A0 ..., Ak and B0, ..., Bk, then the polynomials multiplying sin ωx in y′ p, y′′ p, ay′′ p + by′ p + cyp and y′′ p + ω2yp can be obtained by replacing A0, ..., Ak by B0, ..., Bk and B0, ..., Bk by −A0, ..., −Ak in the corresponding polynomials multiplying cos ωx. We won’t use this theorem in our examples, but we recommend that you use it to check your manipu- lations when you work the exercises. Example 5.5.4 Find a particular solution of y′′ + y = (8 −4x) cos x −(8 + 8x) sin x. (5.5.12) Solution According to Theorem 5.5.1, we should look for a particular solution of the form yp = (A0x + A1x2) cos x + (B0x + B1x2) sin x, (5.5.13) since cos x and sin x are solutions of the complementary equation. However, let’s try yp = (A0 + A1x) cos x + (B0 + B1x) sin x (5.5.14) first, so you can see why it doesn’t work. From (5.5.10), y′′ p = (2B1 −A0 −A1x) cos x −(2A1 + B0 + B1x) sinx, which together with (5.5.14) implies that y′′ p + yp = 2B1 cos x −2A1 sin x. Since the right side of this equation does not contain x cos x or x sinx, (5.5.14) can’t satisfy (5.5.12) no matter how we choose A0, A1, B0, and B1. Now let yp be as in (5.5.13). Then y′ p = A0 + (2A1 + B0)x + B1x2 cos x + B0 + (2B1 −A0)x −A1x2 sin x and y′′ p = 2A1 + 2B0 −(A0 −4B1)x −A1x2 cos x + 2B1 −2A0 −(B0 + 4A1)x −B1x2 sin x, | Elementary Differential Equations with Boundary Value Problems_Page_251_Chunk2656 |
242 Chapter 5 Linear Second Order Equations so y′′ p + yp = (2A1 + 2B0 + 4B1x) cos x + (2B1 −2A0 −4A1x) sin x. Comparing the coefficients of cos x and sin x here with the corresponding coefficients in (5.5.12) shows that yp is a solution of (5.5.12) if 4B1 = −4 −4A1 = −8 2B0 + 2A1 = 8 −2A0 + 2B1 = −8. The solution of this system is A1 = 2, B1 = −1, A0 = 3, B0 = 2. Therefore yp = x [(3 + 2x) cos x + (2 −x) sin x] is a particular solution of (5.5.12). Forcing Functions with Exponential Factors To find a particular solution of ay′′ + by′ + cy = eλx (P (x) cos ωx + Q(x) sinωx) (5.5.15) when λ ̸= 0, we recall from Section 5.4 that substituting y = ueλx into (5.5.15) will produce a constant coefficient equation for u with the forcing function P (x) cosωx + Q(x) sinωx. We can find a particular solution up of this equation by the procedure that we used in Examples 5.5.1–5.5.4. Then yp = upeλx is a particular solution of (5.5.15). Example 5.5.5 Find a particular solution of y′′ −3y′ + 2y = e−2x [2 cos 3x −(34 −150x) sin3x]. (5.5.16) Solution Let y = ue−2x. Then y′′ −3y′ + 2y = e−2x [(u′′ −4u′ + 4u) −3(u′ −2u) + 2u] = e−2x(u′′ −7u′ + 12u) = e−2x [2 cos 3x −(34 −150x) sin3x] if u′′ −7u′ + 12u = 2 cos 3x −(34 −150x) sin3x. (5.5.17) Since cos 3x and sin 3x aren’t solutions of the complementary equation u′′ −7u′ + 12u = 0, Theorem 5.5.1 tells us to look for a particular solution of (5.5.17) of the form up = (A0 + A1x) cos 3x + (B0 + B1x) sin 3x. (5.5.18) Then u′ p = (A1 + 3B0 + 3B1x) cos 3x + (B1 −3A0 −3A1x) sin3x and u′′ p = (−9A0 + 6B1 −9A1x) cos 3x −(9B0 + 6A1 + 9B1x) sin 3x, | Elementary Differential Equations with Boundary Value Problems_Page_252_Chunk2657 |
Section 5.5 The Method of Undetermined Coefficients II 243 so u′′ p −7u′ p + 12up = [3A0 −21B0 −7A1 + 6B1 + (3A1 −21B1)x] cos 3x + [21A0 + 3B0 −6A1 −7B1 + (21A1 + 3B1)x] sin 3x. Comparing the coefficients of x cos 3x, x sin 3x, cos 3x, and sin 3x here with the corresponding coeffi- cients on the right side of (5.5.17) shows that up is a solution of (5.5.17) if 3A1 −21B1 = 0 21A1 + 3B1 = 150 3A0 −21B0 −7A1 + 6B1 = 2 21A0 + 3B0 −6A1 −7B1 = −34. (5.5.19) Solving the first two equations yields A1 = 7, B1 = 1. Substituting these values into the last two equations of (5.5.19) yields 3A0 −21B0 = 2 + 7A1 −6B1 = 45 21A0 + 3B0 = −34 + 6A1 + 7B1 = 15. Solving this system yields A0 = 1, B0 = −2. Substituting A0 = 1, A1 = 7, B0 = −2, and B1 = 1 into (5.5.18) shows that up = (1 + 7x) cos 3x −(2 −x) sin 3x is a particular solution of (5.5.17). Therefore yp = e−2x [(1 + 7x) cos 3x −(2 −x) sin 3x] is a particular solution of (5.5.16). Example 5.5.6 Find a particular solution of y′′ + 2y′ + 5y = e−x [(6 −16x) cos 2x −(8 + 8x) sin2x] . (5.5.20) Solution Let y = ue−x. Then y′′ + 2y′ + 5y = e−x [(u′′ −2u′ + u) + 2(u′ −u) + 5u] = e−x(u′′ + 4u) = e−x [(6 −16x) cos 2x −(8 + 8x) sin2x] if u′′ + 4u = (6 −16x) cos2x −(8 + 8x) sin 2x. (5.5.21) Since cos 2x and sin 2x are solutions of the complementary equation u′′ + 4u = 0, Theorem 5.5.1 tells us to look for a particular solution of (5.5.21) of the form up = (A0x + A1x2) cos 2x + (B0x + B1x2) sin 2x. | Elementary Differential Equations with Boundary Value Problems_Page_253_Chunk2658 |
244 Chapter 5 Linear Second Order Equations Then u′ p = A0 + (2A1 + 2B0)x + 2B1x2 cos 2x + B0 + (2B1 −2A0)x −2A1x2 sin 2x and u′′ p = 2A1 + 4B0 −(4A0 −8B1)x −4A1x2 cos 2x + 2B1 −4A0 −(4B0 + 8A1)x −4B1x2 sin 2x, so u′′ p + 4up = (2A1 + 4B0 + 8B1x) cos 2x + (2B1 −4A0 −8A1x) sin 2x. Equating the coefficients of x cos 2x, x sin 2x, cos 2x, and sin 2x here with the corresponding coefficients on the right side of (5.5.21) shows that up is a solution of (5.5.21) if 8B1 = −16 −8A1 = −8 4B0 + 2A1 = 6 −4A0 + 2B1 = −8. (5.5.22) The solution of this system is A1 = 1, B1 = −2, B0 = 1, A0 = 1. Therefore up = x[(1 + x) cos 2x + (1 −2x) sin 2x] is a particular solution of (5.5.21), and yp = xe−x [(1 + x) cos 2x + (1 −2x) sin 2x] is a particular solution of (5.5.20). You can also find a particular solution of (5.5.20) by substituting yp = xe−x [(A0 + A1x) cos 2x + (B0 + B1x) sin2x] for y in (5.5.20) and equating the coefficients of xe−x cos 2x, xe−x sin 2x, e−x cos 2x, and e−x sin 2x in the resulting expression for y′′ p + 2y′ p + 5yp with the corresponding coefficients on the right side of (5.5.20). (See Exercise 38). This leads to the same system (5.5.22) of equations for A0, A1, B0, and B1 that we obtained in Example 5.5.6. However, if you try this approach you’ll see that deriving (5.5.22) this way is much more tedious than the way we did it in Example 5.5.6. 5.5 Exercises In Exercises 1–17 find a particular solution. 1. y′′ + 3y′ + 2y = 7 cos x −sin x 2. y′′ + 3y′ + y = (2 −6x) cos x −9 sinx 3. y′′ + 2y′ + y = ex(6 cos x + 17 sinx) 4. y′′ + 3y′ −2y = −e2x(5 cos 2x + 9 sin2x) 5. y′′ −y′ + y = ex(2 + x) sinx 6. y′′ + 3y′ −2y = e−2x [(4 + 20x) cos 3x + (26 −32x) sin 3x] | Elementary Differential Equations with Boundary Value Problems_Page_254_Chunk2659 |
Section 5.5 The Method of Undetermined Coefficients II 245 7. y′′ + 4y = −12 cos 2x −4 sin2x 8. y′′ + y = (−4 + 8x) cos x + (8 −4x) sinx 9. 4y′′ + y = −4 cos x/2 −8x sinx/2 10. y′′ + 2y′ + 2y = e−x(8 cos x −6 sin x) 11. y′′ −2y′ + 5y = ex [(6 + 8x) cos 2x + (6 −8x) sin 2x] 12. y′′ + 2y′ + y = 8x2 cos x −4x sinx 13. y′′ + 3y′ + 2y = (12 + 20x + 10x2) cos x + 8x sinx 14. y′′ + 3y′ + 2y = (1 −x −4x2) cos 2x −(1 + 7x + 2x2) sin 2x 15. y′′ −5y′ + 6y = −ex (4 + 6x −x2) cos x −(2 −4x + 3x2) sin x 16. y′′ −2y′ + y = −ex (3 + 4x −x2) cos x + (3 −4x −x2) sin x 17. y′′ −2y′ + 2y = ex (2 −2x −6x2) cos x + (2 −10x + 6x2) sin x In Exercises 1–17 find a particular solution and graph it. 18. C/G y′′ + 2y′ + y = e−x [(5 −2x) cos x −(3 + 3x) sinx] 19. C/G y′′ + 9y = −6 cos 3x −12 sin3x 20. C/G y′′ + 3y′ + 2y = (1 −x −4x2) cos 2x −(1 + 7x + 2x2) sin 2x 21. C/G y′′ + 4y′ + 3y = e−x (2 + x + x2) cos x + (5 + 4x + 2x2) sin x In Exercises 22–26 solve the initial value problem. 22. y′′ −7y′ + 6y = −ex(17 cos x −7 sin x), y(0) = 4, y′(0) = 2 23. y′′ −2y′ + 2y = −ex(6 cos x + 4 sinx), y(0) = 1, y′(0) = 4 24. y′′ + 6y′ + 10y = −40ex sin x, y(0) = 2, y′(0) = −3 25. y′′ −6y′ + 10y = −e3x(6 cos x + 4 sinx), y(0) = 2, y′(0) = 7 26. y′′ −3y′ + 2y = e3x [21 cosx −(11 + 10x) sinx] , y(0) = 0, y′(0) = 6 In Exercises 27–32 use the principle of superposition to find a particular solution. Where indicated, solve the initial value problem. 27. y′′ −2y′ −3y = 4e3x + ex(cos x −2 sinx) 28. y′′ + y = 4 cos x −2 sin x + xex + e−x 29. y′′ −3y′ + 2y = xex + 2e2x + sin x 30. y′′ −2y′ + 2y = 4xex cos x + xe−x + 1 + x2 31. y′′ −4y′ + 4y = e2x(1 + x) + e2x(cos x −sin x) + 3e3x + 1 + x 32. y′′ −4y′ + 4y = 6e2x + 25 sinx, y(0) = 5, y′(0) = 3 In Exercises 33–35 solve the initial value problem and graph the solution. 33. C/G y′′ + 4y = −e−2x [(4 −7x) cos x + (2 −4x) sin x] , y(0) = 3, y′(0) = 1 34. C/G y′′ + 4y′ + 4y = 2 cos 2x + 3 sin 2x + e−x, y(0) = −1, y′(0) = 2 | Elementary Differential Equations with Boundary Value Problems_Page_255_Chunk2660 |
246 Chapter 5 Linear Second Order Equations 35. C/G y′′ + 4y = ex(11 + 15x) + 8 cos 2x −12 sin2x, y(0) = 3, y′(0) = 5 36. (a) Verify that if yp = A(x) cos ωx + B(x) sin ωx where A and B are twice differentiable, then y′ p = (A′ + ωB) cos ωx + (B′ −ωA) sin ωx and y′′ p = (A′′ + 2ωB′ −ω2A) cos ωx + (B′′ −2ωA′ −ω2B) sin ωx. (b) Use the results of (a) to verify that ay′′ p + by′ p + cyp = (c −aω2)A + bωB + 2aωB′ + bA′ + aA′′ cos ωx + −bωA + (c −aω2)B −2aωA′ + bB′ + aB′′ sin ωx. (c) Use the results of (a) to verify that y′′ p + ω2yp = (A′′ + 2ωB′) cos ωx + (B′′ −2ωA′) sin ωx. (d) Prove Theorem 5.5.2. 37. Let a, b, c, and ω be constants, with a ̸= 0 and ω > 0, and let P (x) = p0 + p1x + · · · + pkxk and Q(x) = q0 + q1x + · · · + qkxk, where at least one of the coefficients pk, qk is nonzero, so k is the larger of the degrees of P and Q. (a) Show that if cos ωx and sin ωx are not solutions of the complementary equation ay′′ + by′ + cy = 0, then there are polynomials A(x) = A0 + A1x + · · · + Akxk and B(x) = B0 + B1x + · · · + Bkxk (A) such that (c −aω2)A + bωB + 2aωB′ + bA′ + aA′′ = P −bωA + (c −aω2)B −2aωA′ + bB′ + aB′′ = Q, where (Ak, Bk), (Ak−1, Bk−1), ...,(A0, B0) can be computed successively by solving the systems (c −aω2)Ak + bωBk = pk −bωAk + (c −aω2)Bk = qk, and, if 1 ≤r ≤k, (c −aω2)Ak−r + bωBk−r = pk−r + · · · −bωAk−r + (c −aω2)Bk−r = qk−r + · · · , where the terms indicated by “· · · ” depend upon the previously computed coefficients with subscripts greater than k −r. Conclude from this and Exercise 36(b) that yp = A(x) cos ωx + B(x) sin ωx (B) is a particular solution of ay′′ + by′ + cy = P (x) cos ωx + Q(x) sin ωx. | Elementary Differential Equations with Boundary Value Problems_Page_256_Chunk2661 |
Section 5.5 The Method of Undetermined Coefficients II 247 (b) Conclude from Exercise 36(c) that the equation a(y′′ + ω2y) = P (x) cosωx + Q(x) sin ωx (C) does not have a solution of the form (B) with A and B as in (A). Then show that there are polynomials A(x) = A0x + A1x2 + · · · + Akxk+1 and B(x) = B0x + B1x2 + · · · + Bkxk+1 such that a(A′′ + 2ωB′) = P a(B′′ −2ωA′) = Q, where the pairs (Ak, Bk), (Ak−1, Bk−1), ..., (A0, B0) can be computed successively as follows: Ak = − qk 2aω(k + 1) Bk = pk 2aω(k + 1), and, if k ≥1, Ak−j = −1 2ω qk−j a(k −j + 1) −(k −j + 2)Bk−j+1 Bk−j = 1 2ω pk−j a(k −j + 1) −(k −j + 2)Ak−j+1 for 1 ≤j ≤k. Conclude that (B) with this choice of the polynomials A and B is a particular solution of (C). 38. Show that Theorem 5.5.1 implies the next theorem: Suppose ω is a positive number and P and Q are polynomials. Let k be the larger of the degrees of P and Q. Then the equation ay′′ + by′ + cy = eλx (P (x) cos ωx + Q(x) sin ωx) has a particular solution yp = eλx (A(x) cos ωx + B(x) sin ωx) , (A) where A(x) = A0 + A1x + · · · + Akxk and B(x) = B0 + B1x + · · · + Bkxk, provided that eλx cos ωx and eλx sin ωx are not solutions of the complementary equation. The equation a y′′ −2λy′ + (λ2 + ω2)y = eλx (P (x) cosωx + Q(x) sin ωx) (for which eλx cos ωx and eλx sin ωx are solutions of the complementary equation) has a partic- ular solution of the form (A), where A(x) = A0x + A1x2 + · · · + Akxk+1 and B(x) = B0x + B1x2 + · · · + Bkxk+1. | Elementary Differential Equations with Boundary Value Problems_Page_257_Chunk2662 |
248 Chapter 5 Linear Second Order Equations 39. This exercise presents a method for evaluating the integral y = Z eλx (P (x) cos ωx + Q(x) sinωx) dx where ω ̸= 0 and P (x) = p0 + p1x + · · · + pkxk, Q(x) = q0 + q1x + · · · + qkxk. (a) Show that y = eλxu, where u′ + λu = P (x) cos ωx + Q(x) sinωx. (A) (b) Show that (A) has a particular solution of the form up = A(x) cos ωx + B(x) sin ωx, where A(x) = A0 + A1x + · · · + Akxk, B(x) = B0 + B1x + · · · + Bkxk, and the pairs of coefficients (Ak, Bk), (Ak−1, Bk−1), ...,(A0, B0) can be computed succes- sively as the solutionsof pairs of equations obtained by equating the coefficients of xr cos ωx and xr sin ωx for r = k, k −1, ..., 0. (c) Conclude that Z eλx (P (x) cos ωx + Q(x) sinωx) dx = eλx (A(x) cos ωx + B(x) sin ωx) + c, where c is a constant of integration. 40. Use the method of Exercise 39 to evaluate the integral. (a) R x2 cos x dx (b) R x2ex cos x dx (c) R xe−x sin 2x dx (d) R x2e−x sin x dx (e) R x3ex sin x dx (f) R ex [x cos x −(1 + 3x) sin x] dx (g) R e−x (1 + x2) cos x + (1 −x2) sin x dx 5.6 REDUCTION OF ORDER In this section we give a method for finding the general solution of P0(x)y′′ + P1(x)y′ + P2(x)y = F (x) (5.6.1) if we know a nontrivial solution y1 of the complementary equation P0(x)y′′ + P1(x)y′ + P2(x)y = 0. (5.6.2) The method is called reduction of order because it reduces the task of solving (5.6.1) to solving a first order equation. Unlike the method of undetermined coefficients, it does not require P0, P1, and P2 to be constants, or F to be of any special form. | Elementary Differential Equations with Boundary Value Problems_Page_258_Chunk2663 |
Section 5.6 Reduction of Order 249 By now you shoudn’t be surprised that we look for solutions of (5.6.1) in the form y = uy1 (5.6.3) where u is to be determined so that y satisfies (5.6.1). Substituting (5.6.3) and y′ = u′y1 + uy′ 1 y′′ = u′′y1 + 2u′y′ 1 + uy′′ 1 into (5.6.1) yields P0(x)(u′′y1 + 2u′y′ 1 + uy′′ 1 ) + P1(x)(u′y1 + uy′ 1) + P2(x)uy1 = F (x). Collecting the coefficients of u, u′, and u′′ yields (P0y1)u′′ + (2P0y′ 1 + P1y1)u′ + (P0y′′ 1 + P1y′ 1 + P2y1)u = F. (5.6.4) However, the coefficient of u is zero, since y1 satisfies (5.6.2). Therefore (5.6.4) reduces to Q0(x)u′′ + Q1(x)u′ = F, (5.6.5) with Q0 = P0y1 and Q1 = 2P0y′ 1 + P1y1. (It isn’t worthwhile to memorize the formulas for Q0 and Q1!) Since (5.6.5) is a linear first order equation in u′, we can solve it for u′ by variation of parameters as in Section 1.2, integrate the solution to obtain u, and then obtain y from (5.6.3). Example 5.6.1 (a) Find the general solution of xy′′ −(2x + 1)y′ + (x + 1)y = x2, (5.6.6) given that y1 = ex is a solution of the complementary equation xy′′ −(2x + 1)y′ + (x + 1)y = 0. (5.6.7) (b) As a byproduct of (a), find a fundamental set of solutions of (5.6.7). SOLUTION(a) If y = uex, then y′ = u′ex + uex and y′′ = u′′ex + 2u′ex + uex, so xy′′ −(2x + 1)y′ + (x + 1)y = x(u′′ex + 2u′ex + uex) −(2x + 1)(u′ex + uex) + (x + 1)uex = (xu′′ −u′)ex. Therefore y = uex is a solution of (5.6.6) if and only if (xu′′ −u′)ex = x2, which is a first order equation in u′. We rewrite it as u′′ −u′ x = xe−x. (5.6.8) | Elementary Differential Equations with Boundary Value Problems_Page_259_Chunk2664 |
250 Chapter 5 Linear Second Order Equations To focus on how we apply variation of parameters to this equation, we temporarily write z = u′, so that (5.6.8) becomes z′ −z x = xe−x. (5.6.9) We leave it to you to show (by separation of variables) that z1 = x is a solution of the complementary equation z′ −z x = 0 for (5.6.9). By applying variation of parameters as in Section 1.2, we can now see that every solution of (5.6.9) is of the form z = vx where v′x = xe−x, so v′ = e−x and v = −e−x + C1. Since u′ = z = vx, u is a solution of (5.6.8) if and only if u′ = vx = −xe−x + C1x. Integrating this yields u = (x + 1)e−x + C1 2 x2 + C2. Therefore the general solution of (5.6.6) is y = uex = x + 1 + C1 2 x2ex + C2ex. (5.6.10) SOLUTION(b) By letting C1 = C2 = 0 in (5.6.10), we see that yp1 = x + 1 is a solution of (5.6.6). By letting C1 = 2 and C2 = 0, we see that yp2 = x+1+x2ex is also a solution of (5.6.6). Since the difference of two solutions of (5.6.6) is a solution of (5.6.7), y2 = yp1 −yp2 = x2ex is a solution of (5.6.7). Since y2/y1 is nonconstant and we already know that y1 = ex is a solution of (5.6.6), Theorem 5.1.6 implies that {ex, x2ex} is a fundamental set of solutions of (5.6.7). Although (5.6.10) is a correct form for the general solution of (5.6.6), it’s silly to leave the arbitrary co- efficient of x2ex as C1/2 where C1 is an arbitrary constant. Moreover, it’s sensible to make the subscripts of the coefficients of y1 = ex and y2 = x2ex consistent with the subscripts of the functions themselves. Therefore we rewrite (5.6.10) as y = x + 1 + c1ex + c2x2ex by simply renaming the arbitrary constants. We’ll also do this in the next two examples, and in the answers to the exercises. Example 5.6.2 (a) Find the general solution of x2y′′ + xy′ −y = x2 + 1, given that y1 = x is a solution of the complementary equation x2y′′ + xy′ −y = 0. (5.6.11) As a byproduct of this result, find a fundamental set of solutions of (5.6.11). (b) Solve the initial value problem x2y′′ + xy′ −y = x2 + 1, y(1) = 2, y′(1) = −3. (5.6.12) | Elementary Differential Equations with Boundary Value Problems_Page_260_Chunk2665 |
Section 5.6 Reduction of Order 251 SOLUTION(a) If y = ux, then y′ = u′x + u and y′′ = u′′x + 2u′, so x2y′′ + xy′ −y = x2(u′′x + 2u′) + x(u′x + u) −ux = x3u′′ + 3x2u′. Therefore y = ux is a solution of (5.6.12) if and only if x3u′′ + 3x2u′ = x2 + 1, which is a first order equation in u′. We rewrite it as u′′ + 3 xu′ = 1 x + 1 x3 . (5.6.13) To focus on how we apply variation of parameters to this equation, we temporarily write z = u′, so that (5.6.13) becomes z′ + 3 xz = 1 x + 1 x3 . (5.6.14) We leave it to you to show by separation of variables that z1 = 1/x3 is a solution of the complementary equation z′ + 3 xz = 0 for (5.6.14). By variation of parameters, every solution of (5.6.14) is of the form z = v x3 where v′ x3 = 1 x + 1 x3 , so v′ = x2 + 1 and v = x3 3 + x + C1. Since u′ = z = v/x3, u is a solution of (5.6.14) if and only if u′ = v x3 = 1 3 + 1 x2 + C1 x3 . Integrating this yields u = x 3 −1 x −C1 2x2 + C2. Therefore the general solution of (5.6.12) is y = ux = x2 3 −1 −C1 2x + C2x. (5.6.15) Reasoning as in the solution of Example 5.6.1(a), we conclude that y1 = x and y2 = 1/x form a fundamental set of solutions for (5.6.11). As we explained above, we rename the constants in (5.6.15) and rewrite it as y = x2 3 −1 + c1x + c2 x . (5.6.16) SOLUTION(b) Differentiating (5.6.16) yields y′ = 2x 3 + c1 −c2 x2 . (5.6.17) | Elementary Differential Equations with Boundary Value Problems_Page_261_Chunk2666 |
252 Chapter 5 Linear Second Order Equations Setting x = 1 in (5.6.16) and (5.6.17) and imposing the initial conditions y(1) = 2 and y′(1) = −3 yields c1 + c2 = 8 3 c1 −c2 = −11 3 . Solving these equations yields c1 = −1/2, c2 = 19/6. Therefore the solution of (5.6.12) is y = x2 3 −1 −x 2 + 19 6x. Using reduction of order to find the general solution of a homogeneous linear second order equation leads to a homogeneous linear first order equation in u′ that can be solved by separation of variables. The next example illustrates this. Example 5.6.3 Find the general solution and a fundamental set of solutions of x2y′′ −3xy′ + 3y = 0, (5.6.18) given that y1 = x is a solution. Solution If y = ux then y′ = u′x + u and y′′ = u′′x + 2u′, so x2y′′ −3xy′ + 3y = x2(u′′x + 2u′) −3x(u′x + u) + 3ux = x3u′′ −x2u′. Therefore y = ux is a solution of (5.6.18) if and only if x3u′′ −x2u′ = 0. Separating the variables u′ and x yields u′′ u′ = 1 x, so ln |u′| = ln|x| + k, or, equivalently, u′ = C1x. Therefore u = C1 2 x2 + C2, so the general solution of (5.6.18) is y = ux = C1 2 x3 + C2x, which we rewrite as y = c1x + c2x3. Therefore {x, x3} is a fundamental set of solutions of (5.6.18). 5.6 Exercises In Exercises 1–17 find the general solution, given that y1 satisfies the complementary equation. As a byproduct, find a fundamental set of solutions of the complementary equation. | Elementary Differential Equations with Boundary Value Problems_Page_262_Chunk2667 |
Section 5.6 Reduction of Order 253 1. (2x + 1)y′′ −2y′ −(2x + 3)y = (2x + 1)2; y1 = e−x 2. x2y′′ + xy′ −y = 4 x2 ; y1 = x 3. x2y′′ −xy′ + y = x; y1 = x 4. y′′ −3y′ + 2y = 1 1 + e−x ; y1 = e2x 5. y′′ −2y′ + y = 7x3/2ex; y1 = ex 6. 4x2y′′ + (4x −8x2)y′ + (4x2 −4x −1)y = 4x1/2ex(1 + 4x); y1 = x1/2ex 7. y′′ −2y′ + 2y = ex sec x; y1 = ex cos x 8. y′′ + 4xy′ + (4x2 + 2)y = 8e−x(x+2); y1 = e−x2 9. x2y′′ + xy′ −4y = −6x −4; y1 = x2 10. x2y′′ + 2x(x −1)y′ + (x2 −2x + 2)y = x3e2x; y1 = xe−x 11. x2y′′ −x(2x −1)y′ + (x2 −x −1)y = x2ex; y1 = xex 12. (1 −2x)y′′ + 2y′ + (2x −3)y = (1 −4x + 4x2)ex; y1 = ex 13. x2y′′ −3xy′ + 4y = 4x4; y1 = x2 14. 2xy′′ + (4x + 1)y′ + (2x + 1)y = 3x1/2e−x; y1 = e−x 15. xy′′ −(2x + 1)y′ + (x + 1)y = −ex; y1 = ex 16. 4x2y′′ −4x(x + 1)y′ + (2x + 3)y = 4x5/2e2x; y1 = x1/2 17. x2y′′ −5xy′ + 8y = 4x2; y1 = x2 In Exercises 18–30 find a fundamental set of solutions, given that y1 is a solution. 18. xy′′ + (2 −2x)y′ + (x −2)y = 0; y1 = ex 19. x2y′′ −4xy′ + 6y = 0; y1 = x2 20. x2(ln |x|)2y′′ −(2x ln|x|)y′ + (2 + ln|x|)y = 0; y1 = ln |x| 21. 4xy′′ + 2y′ + y = 0; y1 = sin √x 22. xy′′ −(2x + 2)y′ + (x + 2)y = 0; y1 = ex 23. x2y′′ −(2a −1)xy′ + a2y = 0; y1 = xa 24. x2y′′ −2xy′ + (x2 + 2)y = 0; y1 = x sinx 25. xy′′ −(4x + 1)y′ + (4x + 2)y = 0; y1 = e2x 26. 4x2(sin x)y′′ −4x(x cos x + sinx)y′ + (2x cos x + 3 sin x)y = 0; y1 = x1/2 27. 4x2y′′ −4xy′ + (3 −16x2)y = 0; y1 = x1/2e2x 28. (2x + 1)xy′′ −2(2x2 −1)y′ −4(x + 1)y = 0; y1 = 1/x 29. (x2 −2x)y′′ + (2 −x2)y′ + (2x −2)y = 0; y1 = ex 30. xy′′ −(4x + 1)y′ + (4x + 2)y = 0; y1 = e2x In Exercises 31–33 solve the initial value problem, given that y1 satisfies the complementary equation. 31. x2y′′ −3xy′ + 4y = 4x4, y(−1) = 7, y′(−1) = −8; y1 = x2 32. (3x −1)y′′ −(3x + 2)y′ −(6x −8)y = 0, y(0) = 2, y′(0) = 3; y1 = e2x | Elementary Differential Equations with Boundary Value Problems_Page_263_Chunk2668 |
254 Chapter 5 Linear Second Order Equations 33. (x + 1)2y′′ −2(x + 1)y′ −(x2 + 2x −1)y = (x + 1)3ex, y(0) = 1, y′(0) = −1; y1 = (x + 1)ex In Exercises 34 and 35 solve the initial value problem and graph the solution, given that y1 satisfies the complementary equation. 34. C/G x2y′′ + 2xy′ −2y = x2, y(1) = 5 4, y′(1) = 3 2; y1 = x 35. C/G (x2 −4)y′′ + 4xy′ + 2y = x + 2, y(0) = −1 3, y′(0) = −1; y1 = 1 x −2 36. Suppose p1 and p2 are continuous on (a, b). Let y1 be a solution of y′′ + p1(x)y′ + p2(x)y = 0 (A) that has no zeros on (a, b), and let x0 be in (a, b). Use reduction of order to show that y1 and y2(x) = y1(x) Z x x0 1 y2 1(t) exp − Z t x0 p1(s) ds dt form a fundamental set of solutionsof (A) on (a, b). (NOTE: This exercise is related to Exercise 9.) 37. The nonlinear first order equation y′ + y2 + p(x)y + q(x) = 0 (A) is a Riccati equation. (See Exercise 2.4.55.) Assume that p and q are continuous. (a) Show that y is a solution of (A) if and only if y = z′/z, where z′′ + p(x)z′ + q(x)z = 0. (B) (b) Show that the general solution of (A) is y = c1z′ 1 + c2z′ 2 c1z1 + c2z2 , (C) where {z1, z2} is a fundamental set of solutions of (B) and c1 and c2 are arbitrary constants. (c) Does the formula (C) imply that the first order equation (A) has a two–parameter family of solutions? Explain your answer. 38. Use a method suggested by Exercise 37 to find all solutions. of the equation. (a) y′ + y2 + k2 = 0 (b) y′ + y2 −3y + 2 = 0 (c) y′ + y2 + 5y −6 = 0 (d) y′ + y2 + 8y + 7 = 0 (e) y′ + y2 + 14y + 50 = 0 (f) 6y′ + 6y2 −y −1 = 0 (g) 36y′ + 36y2 −12y + 1 = 0 39. Use a method suggested by Exercise 37 and reduction of order to find all solutions of the equation, given that y1 is a solution. (a) x2(y′ + y2) −x(x + 2)y + x + 2 = 0; y1 = 1/x (b) y′ + y2 + 4xy + 4x2 + 2 = 0; y1 = −2x (c) (2x + 1)(y′ + y2) −2y −(2x + 3) = 0; y1 = −1 (d) (3x −1)(y′ + y2) −(3x + 2)y −6x + 8 = 0; y1 = 2 | Elementary Differential Equations with Boundary Value Problems_Page_264_Chunk2669 |
Section 5.7 Variation of Parameters 255 (e) x2(y′ + y2) + xy + x2 −1 4 = 0; y1 = −tan x −1 2x (f) x2(y′ + y2) −7xy + 7 = 0; y1 = 1/x 40. The nonlinear first order equation y′ + r(x)y2 + p(x)y + q(x) = 0 (A) is the generalized Riccati equation. (See Exercise 2.4.55.) Assume that p and q are continuous and r is differentiable. (a) Show that y is a solution of (A) if and only if y = z′/rz, where z′′ + p(x) −r′(x) r(x) z′ + r(x)q(x)z = 0. (B) (b) Show that the general solution of (A) is y = c1z′ 1 + c2z′ 2 r(c1z1 + c2z2), where {z1, z2} is a fundamental set of solutions of (B) and c1 and c2 are arbitrary constants. 5.7 VARIATION OF PARAMETERS In this section we give a method called variation of parameters for finding a particular solution of P0(x)y′′ + P1(x)y′ + P2(x)y = F (x) (5.7.1) if we know a fundamental set {y1, y2} of solutions of the complementary equation P0(x)y′′ + P1(x)y′ + P2(x)y = 0. (5.7.2) Having found a particular solution yp by this method, we can write the general solution of (5.7.1) as y = yp + c1y1 + c2y2. Since we need only one nontrivial solution of (5.7.2) to find the general solution of (5.7.1) by reduction of order, it’s natural to ask why we’re interested in variation of parameters, which requires two linearly independent solutions of (5.7.2) to achieve the same goal. Here’s the answer: • If we already know two linearly independent solutions of (5.7.2) then variation of parameters will probably be simpler than reduction of order. • Variation of parameters generalizes naturally to a method for finding particular solutions of higher order linear equations (Section 9.4) and linear systems of equations (Section 10.7), while reduction of order doesn’t. • Variation of parameters is a powerful theoretical tool used by researchers in differential equations. Although a detailed discussion of this is beyond the scope of this book, you can get an idea of what it means from Exercises 37–39. | Elementary Differential Equations with Boundary Value Problems_Page_265_Chunk2670 |
256 Chapter 5 Linear Second Order Equations We’ll now derive the method. As usual, we consider solutions of (5.7.1) and (5.7.2) on an interval (a, b) where P0, P1, P2, and F are continuous and P0 has no zeros. Suppose that {y1, y2} is a fundamental set of solutions of the complementary equation (5.7.2). We look for a particular solution of (5.7.1) in the form yp = u1y1 + u2y2 (5.7.3) where u1 and u2 are functions to be determined so that yp satisfies (5.7.1). You may not think this is a good idea, since there are now two unknown functions to be determined, rather than one. However, since u1 and u2 have to satisfy only one condition (that yp is a solution of (5.7.1)), we can impose a second condition that produces a convenient simplification, as follows. Differentiating (5.7.3) yields y′ p = u1y′ 1 + u2y′ 2 + u′ 1y1 + u′ 2y2. (5.7.4) As our second condition on u1 and u2 we require that u′ 1y1 + u′ 2y2 = 0. (5.7.5) Then (5.7.4) becomes y′ p = u1y′ 1 + u2y′ 2; (5.7.6) that is, (5.7.5) permits us to differentiate yp (once!) as if u1 and u2 are constants. Differentiating (5.7.4) yields y′′ p = u1y′′ 1 + u2y′′ 2 + u′ 1y′ 1 + u′ 2y′ 2. (5.7.7) (There are no terms involving u′′ 1 and u′′ 2 here, as there would be if we hadn’t required (5.7.5).) Substitut- ing (5.7.3), (5.7.6), and (5.7.7) into (5.7.1) and collecting the coefficients of u1 and u2 yields u1(P0y′′ 1 + P1y′ 1 + P2y1) + u2(P0y′′ 2 + P1y′ 2 + P2y2) + P0(u′ 1y′ 1 + u′ 2y′ 2) = F. As in the derivation of the method of reduction of order, the coefficients of u1 and u2 here are both zero because y1 and y2 satisfy the complementary equation. Hence, we can rewrite the last equation as P0(u′ 1y′ 1 + u′ 2y′ 2) = F. (5.7.8) Therefore yp in (5.7.3) satisfies (5.7.1) if u′ 1y1 + u′ 2y2 = 0 u′ 1y′ 1 + u′ 2y′ 2 = F P0 , (5.7.9) where the first equation is the same as (5.7.5) and the second is from (5.7.8). We’ll now show that you can always solve (5.7.9) for u′ 1 and u′ 2. (The method that we use here will always work, but simpler methods usually work when you’re dealing with specific equations.) To obtain u′ 1, multiply the first equation in (5.7.9) by y′ 2 and the second equation by y2. This yields u′ 1y1y′ 2 + u′ 2y2y′ 2 = 0 u′ 1y′ 1y2 + u′ 2y′ 2y2 = F y2 P0 . Subtracting the second equation from the first yields u′ 1(y1y′ 2 −y′ 1y2) = −F y2 P0 . (5.7.10) | Elementary Differential Equations with Boundary Value Problems_Page_266_Chunk2671 |
Section 5.7 Variation of Parameters 257 Since {y1, y2} is a fundamental set of solutions of (5.7.2) on (a, b), Theorem 5.1.6 implies that the Wronskian y1y′ 2 −y′ 1y2 has no zeros on (a, b). Therefore we can solve (5.7.10) for u′ 1, to obtain u′ 1 = − F y2 P0(y1y′ 2 −y′ 1y2). (5.7.11) We leave it to you to start from (5.7.9) and show by a similar argument that u′ 2 = F y1 P0(y1y′ 2 −y′ 1y2). (5.7.12) We can now obtain u1 and u2 by integrating u′ 1 and u′ 2. The constants of integration can be taken to be zero, since any choice of u1 and u2 in (5.7.3) will suffice. You should not memorize (5.7.11) and (5.7.12). On the other hand, you don’t want to rederive the whole procedure for every specific problem. We recommend the a compromise: (a) Write yp = u1y1 + u2y2 (5.7.13) to remind yourself of what you’re doing. (b) Write the system u′ 1y1 + u′ 2y2 = 0 u′ 1y′ 1 + u′ 2y′ 2 = F P0 (5.7.14) for the specific problem you’re trying to solve. (c) Solve (5.7.14) for u′ 1 and u′ 2 by any convenient method. (d) Obtain u1 and u2 by integrating u′ 1 and u′ 2, taking the constants of integration to be zero. (e) Substitute u1 and u2 into (5.7.13) to obtain yp. Example 5.7.1 Find a particular solution yp of x2y′′ −2xy′ + 2y = x9/2, (5.7.15) given that y1 = x and y2 = x2 are solutions of the complementary equation x2y′′ −2xy′ + 2y = 0. Then find the general solution of (5.7.15). Solution We set yp = u1x + u2x2, where u′ 1x + u′ 2x2 = 0 u′ 1 + 2u′ 2x = x9/2 x2 = x5/2. From the first equation, u′ 1 = −u′ 2x. Substituting this into the second equation yields u′ 2x = x5/2, so u′ 2 = x3/2 and therefore u′ 1 = −u′ 2x = −x5/2. Integrating and taking the constants of integration to be zero yields u1 = −2 7x7/2 and u2 = 2 5x5/2. | Elementary Differential Equations with Boundary Value Problems_Page_267_Chunk2672 |
258 Chapter 5 Linear Second Order Equations Therefore yp = u1x + u2x2 = −2 7x7/2x + 2 5x5/2x2 = 4 35x9/2, and the general solution of (5.7.15) is y = 4 35x9/2 + c1x + c2x2. Example 5.7.2 Find a particular solution yp of (x −1)y′′ −xy′ + y = (x −1)2, (5.7.16) given that y1 = x and y2 = ex are solutions of the complementary equation (x −1)y′′ −xy′ + y = 0. Then find the general solution of (5.7.16). Solution We set yp = u1x + u2ex, where u′ 1x + u′ 2ex = 0 u′ 1 + u′ 2ex = (x −1)2 x −1 = x −1. Subtracting the first equation from the second yields −u′ 1(x −1) = x −1, so u′ 1 = −1. From this and the first equation, u′ 2 = −xe−xu′ 1 = xe−x. Integrating and taking the constants of integration to be zero yields u1 = −x and u2 = −(x + 1)e−x. Therefore yp = u1x + u2ex = (−x)x + (−(x + 1)e−x)ex = −x2 −x −1, so the general solution of (5.7.16) is y = yp + c1x + c2ex = −x2 −x −1 + c1x + c2ex = −x2 −1 + (c1 −1)x + c2ex. (5.7.17) However, since c1 is an arbitrary constant, so is c1−1; therefore, we improve the appearance of this result by renaming the constant and writing the general solution as y = −x2 −1 + c1x + c2ex. (5.7.18) There’s nothing wrong with leaving the general solution of (5.7.16) in the form (5.7.17); however, we think you’ll agree that (5.7.18) is preferable. We can also view the transition from (5.7.17) to (5.7.18) differently. In this example the particular solutionyp = −x2−x−1 contained the term −x, which satisfies the complementary equation. We can drop this term and redefine yp = −x2 −1, since −x2 −x −1 is a solution of (5.7.16) and x is a solutionof the complementary equation; hence, −x2−1 = (−x2−x−1)+x is also a solution of (5.7.16). In general, it’s always legitimate to drop linear combinations of {y1, y2} from particular solutions obtained by variation of parameters. (See Exercise 36 for a general discussion of this question.) We’ll do this in the following examples and in the answers to exercises that ask for a particular solution. Therefore, don’t be concerned if your answer to such an exercise differs from ours only by a solution of the complementary equation. | Elementary Differential Equations with Boundary Value Problems_Page_268_Chunk2673 |
Section 5.7 Variation of Parameters 259 Example 5.7.3 Find a particular solution of y′′ + 3y′ + 2y = 1 1 + ex . (5.7.19) Then find the general solution. Solution The characteristic polynomial of the complementary equation y′′ + 3y′ + 2y = 0 (5.7.20) is p(r) = r2 + 3r + 2 = (r + 1)(r + 2), so y1 = e−x and y2 = e−2x form a fundamental set of solutions of (5.7.20). We look for a particular solution of (5.7.19) in the form yp = u1e−x + u2e−2x, where u′ 1e−x + u′ 2e−2x = 0 −u′ 1e−x −2u′ 2e−2x = 1 1 + ex . Adding these two equations yields −u′ 2e−2x = 1 1 + ex , so u′ 2 = −e2x 1 + ex . From the first equation, u′ 1 = −u′ 2e−x = ex 1 + ex . Integrating by means of the substitution v = ex and taking the constants of integration to be zero yields u1 = Z ex 1 + ex dx = Z dv 1 + v = ln(1 + v) = ln(1 + ex) and u2 = − Z e2x 1 + ex dx = − Z v 1 + v dv = Z 1 1 + v −1 dv = ln(1 + v) −v = ln(1 + ex) −ex. Therefore yp = u1e−x + u2e−2x = [ln(1 + ex)]e−x + [ln(1 + ex) −ex] e−2x, so yp = | Elementary Differential Equations with Boundary Value Problems_Page_269_Chunk2674 |
260 Chapter 5 Linear Second Order Equations Example 5.7.4 Solve the initial value problem (x2 −1)y′′ + 4xy′ + 2y = 2 x + 1, y(0) = −1, y′(0) = −5, (5.7.21) given that y1 = 1 x −1 and y2 = 1 x + 1 are solutions of the complementary equation (x2 −1)y′′ + 4xy′ + 2y = 0. Solution We first use variation of parameters to find a particular solution of (x2 −1)y′′ + 4xy′ + 2y = 2 x + 1 on (−1, 1) in the form yp = u1 x −1 + u2 x + 1, where u′ 1 x −1 + u′ 2 x + 1 = 0 (5.7.22) − u′ 1 (x −1)2 − u′ 2 (x + 1)2 = 2 (x + 1)(x2 −1). Multiplying the first equation by 1/(x −1) and adding the result to the second equation yields 1 x2 −1 − 1 (x + 1)2 u′ 2 = 2 (x + 1)(x2 −1). (5.7.23) Since 1 x2 −1 − 1 (x + 1)2 = (x + 1) −(x −1) (x + 1)(x2 −1) = 2 (x + 1)(x2 −1), (5.7.23) implies that u′ 2 = 1. From (5.7.22), u′ 1 = −x −1 x + 1u′ 2 = −x −1 x + 1. Integrating and taking the constants of integration to be zero yields u1 = − Z x −1 x + 1 dx = − Z x + 1 −2 x + 1 dx = Z 2 x + 1 −1 dx = 2 ln(x + 1) −x and u2 = Z dx = x. | Elementary Differential Equations with Boundary Value Problems_Page_270_Chunk2675 |
Section 5.7 Variation of Parameters 261 Therefore yp = u1 x −1 + u2 x + 1 = [2 ln(x + 1) −x] 1 x −1 + x 1 x + 1 = 2 ln(x + 1) x −1 + x 1 x + 1 − 1 x −1 = 2 ln(x + 1) x −1 − 2x (x + 1)(x −1). However, since 2x (x + 1)(x −1) = 1 x + 1 + 1 x −1 is a solution of the complementary equation, we redefine yp = 2 ln(x + 1) x −1 . Therefore the general solution of (5.7.24) is y = 2 ln(x + 1) x −1 + c1 x −1 + c2 x + 1. (5.7.24) Differentiating this yields y′ = 2 x2 −1 −2 ln(x + 1) (x −1)2 − c1 (x −1)2 − c2 (x + 1)2 . Setting x = 0 in the last two equations and imposing the initial conditions y(0) = −1 and y′(0) = −5 yields the system −c1 + c2 = −1 −2 −c1 −c2 = −5. The solution of this system is c1 = 2, c2 = 1. Substituting these into (5.7.24) yields y = 2 ln(x + 1) x −1 + 2 x −1 + 1 x + 1 = 2 ln(x + 1) x −1 + 3x + 1 x2 −1 as the solution of (5.7.21). Figure 5.7.1 is a graph of the solution. Comparison of Methods We’ve now considered three methods for solving nonhomogeneous linear equations: undetermined co- efficients, reduction of order, and variation of parameters. It’s natural to ask which method is best for a given problem. The method of undetermined coefficients should be used for constant coefficient equa- tions with forcing functions that are linear combinations of polynomials multiplied by functions of the form eαx, eλx cos ωx, or eλx sin ωx. Although the other two methods can be used to solve such problems, they will be more difficult except in the most trivial cases, because of the integrations involved. If the equation isn’t a constant coefficient equation or the forcing function isn’t of the form just spec- ified, the method of undetermined coefficients does not apply and the choice is necessarily between the other two methods. The case could be made that reduction of order is better because it requires only one solution of the complementary equation while variation of parameters requires two. However, vari- ation of parameters will probably be easier if you already know a fundamental set of solutions of the complementary equation. | Elementary Differential Equations with Boundary Value Problems_Page_271_Chunk2676 |
262 Chapter 5 Linear Second Order Equations 1.0 −1.0 0.5 0.5 −0.5 10 20 30 40 50 −10 −20 −30 −40 −50 x y Figure 5.7.1 y = 2 ln(x + 1) x −1 + 3x + 1 x2 −1 5.7 Exercises In Exercises 1–6 use variation of parameters to find a particular solution. 1. y′′ + 9y = tan 3x 2. y′′ + 4y = sin 2x sec2 2x 3. y′′ −3y′ + 2y = 4 1 + e−x 4. y′′ −2y′ + 2y = 3ex sec x 5. y′′ −2y′ + y = 14x3/2ex 6. y′′ −y = 4e−x 1 −e−2x | Elementary Differential Equations with Boundary Value Problems_Page_272_Chunk2677 |
Section 5.7 Variation of Parameters 263 In Exercises 7–29 use variation of parameters to find a particular solution, given the solutions y1, y2 of the complementary equation. 7. x2y′′ + xy′ −y = 2x2 + 2; y1 = x, y2 = 1 x 8. xy′′ + (2 −2x)y′ + (x −2)y = e2x; y1 = ex, y2 = ex x 9. 4x2y′′ + (4x −8x2)y′ + (4x2 −4x −1)y = 4x1/2ex, x > 0; y1 = x1/2ex, y2 = x−1/2ex 10. y′′ + 4xy′ + (4x2 + 2)y = 4e−x(x+2); y1 = e−x2, y2 = xe−x2 11. x2y′′ −4xy′ + 6y = x5/2, x > 0; y1 = x2, y2 = x3 12. x2y′′ −3xy′ + 3y = 2x4 sin x; y1 = x, y2 = x3 13. (2x + 1)y′′ −2y′ −(2x + 3)y = (2x + 1)2e−x; y1 = e−x, y2 = xex 14. 4xy′′ + 2y′ + y = sin √x; y1 = cos √x, y2 = sin√x 15. xy′′ −(2x + 2)y′ + (x + 2)y = 6x3ex; y1 = ex, y2 = x3ex 16. x2y′′ −(2a −1)xy′ + a2y = xa+1; y1 = xa, y2 = xa lnx 17. x2y′′ −2xy′ + (x2 + 2)y = x3 cos x; y1 = x cos x, y2 = x sinx 18. xy′′ −y′ −4x3y = 8x5; y1 = ex2, y2 = e−x2 19. (sin x)y′′ + (2 sin x −cos x)y′ + (sin x −cos x)y = e−x; y1 = e−x, y2 = e−x cos x 20. 4x2y′′ −4xy′ + (3 −16x2)y = 8x5/2; y1 = √xe2x, y2 = √xe−2x 21. 4x2y′′ −4xy′ + (4x2 + 3)y = x7/2; y1 = √x sin x, y2 = √x cos x 22. x2y′′ −2xy′ −(x2 −2)y = 3x4; y1 = xex, y2 = xe−x 23. x2y′′ −2x(x + 1)y′ + (x2 + 2x + 2)y = x3ex; y1 = xex, y2 = x2ex 24. x2y′′ −xy′ −3y = x3/2; y1 = 1/x, y2 = x3 25. x2y′′ −x(x + 4)y′ + 2(x + 3)y = x4ex; y1 = x2, y2 = x2ex 26. x2y′′ −2x(x + 2)y′ + (x2 + 4x + 6)y = 2xex; y1 = x2ex, y2 = x3ex 27. x2y′′ −4xy′ + (x2 + 6)y = x4; y1 = x2 cos x, y2 = x2 sin x 28. (x −1)y′′ −xy′ + y = 2(x −1)2ex; y1 = x, y2 = ex 29. 4x2y′′ −4x(x + 1)y′ + (2x + 3)y = x5/2ex; y1 = √x, y2 = √xex In Exercises 30–32 use variation of parameters to solve the initial value problem, given y1, y2 are solu- tions of the complementary equation. 30. (3x −1)y′′ −(3x + 2)y′ −(6x −8)y = (3x −1)2e2x, y(0) = 1, y′(0) = 2; y1 = e2x, y2 = xe−x 31. (x −1)2y′′ −2(x −1)y′ + 2y = (x −1)2, y(0) = 3, y′(0) = −6; y1 = x −1, y2 = x2 −1 32. (x −1)2y′′ −(x2 −1)y′ + (x + 1)y = (x −1)3ex, y(0) = 4, y′(0) = −6; y1 = (x −1)ex, y2 = x −1 | Elementary Differential Equations with Boundary Value Problems_Page_273_Chunk2678 |
264 Chapter 5 Linear Second Order Equations In Exercises 33–35 use variation of parameters to solve the initial value problem and graph the solution, given that y1, y2 are solutions of the complementary equation. 33. C/G (x2 −1)y′′ + 4xy′ + 2y = 2x, y(0) = 0, y′(0) = −2; y1 = 1 x −1, y2 = 1 x + 1 34. C/G x2y′′ + 2xy′ −2y = −2x2, y(1) = 1, y′(1) = −1; y1 = x, y2 = 1 x2 35. C/G (x + 1)(2x + 3)y′′ + 2(x + 2)y′ −2y = (2x + 3)2, y(0) = 0, y′(0) = 0; y1 = x + 2, y2 = 1 x + 1 36. Suppose yp = y + a1y1 + a2y2 is a particular solution of P0(x)y′′ + P1(x)y′ + P2(x)y = F (x), (A) where y1 and y2 are solutions of the complementary equation P0(x)y′′ + P1(x)y′ + P2(x)y = 0. Show that y is also a solution of (A). 37. Suppose p, q, and f are continuous on (a, b) and let x0 be in (a, b). Let y1 and y2 be the solutions of y′′ + p(x)y′ + q(x)y = 0 such that y1(x0) = 1, y′ 1(x0) = 0, y2(x0) = 0, y′ 2(x0) = 1. Use variation of parameters to show that the solution of the initial value problem y′′ + p(x)y′ + q(x)y = f(x), y(x0) = k0, y′(x0) = k1, is y(x) = k0y1(x) + k1y2(x) + Z x x0 (y1(t)y2(x) −y1(x)y2(t)) f(t) exp Z t x0 p(s) ds dt. HINT: Use Abel’s formula for the Wronskian of {y1, y2}, and integrate u′ 1 and u′ 2 from x0 to x. Show also that y′(x) = k0y′ 1(x) + k1y′ 2(x) + Z x x0 (y1(t)y′ 2(x) −y′ 1(x)y2(t)) f(t) exp Z t x0 p(s) ds dt. 38. Suppose f is continuous on an open interval that contains x0 = 0. Use variation of parameters to find a formula for the solution of the initial value problem y′′ −y = f(x), y(0) = k0, y′(0) = k1. 39. Suppose f is continuous on (a, ∞), where a < 0, so x0 = 0 is in (a, ∞). | Elementary Differential Equations with Boundary Value Problems_Page_274_Chunk2679 |
Section 5.7 Variation of Parameters 265 (a) Use variation of parameters to find a formula for the solution of the initial value problem y′′ + y = f(x), y(0) = k0, y′(0) = k1. HINT: You will need the addition formulas for the sine and cosine: sin(A + B) = sin A cos B + cos A sin B cos(A + B) = cos A cos B −sin A sin B. For the rest of this exercise assume that the improper integral R ∞ 0 f(t) dt is absolutely convergent. (b) Show that if y is a solution of y′′ + y = f(x) (A) on (a, ∞), then lim x→∞(y(x) −A0 cos x −A1 sin x) = 0 (B) and lim x→∞(y′(x) + A0 sin x −A1 cos x) = 0, (C) where A0 = k0 − Z ∞ 0 f(t) sin t dt and A1 = k1 + Z ∞ 0 f(t) cos t dt. HINT: Recall from calculus that if R ∞ 0 f(t) dt converges absolutely, then limx→∞ R ∞ x |f(t)| dt = 0. (c) Show that if A0 and A1 are arbitrary constants, then there’s a unique solution of y′′ + y = f(x) on (a, ∞) that satisfies (B) and (C). | Elementary Differential Equations with Boundary Value Problems_Page_275_Chunk2680 |
CHAPTER 6 Applications of Linear Second Order Equations IN THIS CHAPTER we study applications of linear second order equations. SECTIONS 6.1 AND 6.2 is about spring–mass systems. SECTION 6.2 is about RLC circuits, the electrical analogs of spring–mass systems. SECTION 6.3 is about motion of an object under a central force, which is particularly relevant in the space age, since, for example, a satellite moving in orbit subject only to Earth’s gravity is experiencing motion under a central force. 267 | Elementary Differential Equations with Boundary Value Problems_Page_277_Chunk2681 |
268 Chapter 6 Applications of Linear Second Order Equations 6.1 SPRING PROBLEMS I We consider the motion of an object of mass m, suspended from a spring of negligible mass. We say that the spring–mass system is in equilibrium when the object is at rest and the forces acting on it sum to zero. The position of the object in this case is the equilibrium position. We define y to be the displacement of the object from its equilibrium position (Figure 6.1.1), measured positive upward. y (a) 0 (b) (c) Figure 6.1.1 (a) y > 0 (b) y = 0, (c) y < 0 Figure 6.1.2 A spring – mass system with damping Our model accounts for the following kinds of forces acting on the object: • The force −mg, due to gravity. • A force Fs exerted by the spring resisting change in its length. The natural length of the spring is its length with no mass attached. We assume that the spring obeys Hooke’s law: If the length of the spring is changed by an amount ∆L from its natural length, then the spring exerts a force Fs = k∆L, where k is a positive number called the spring constant. If the spring is stretched then ∆L > 0 and Fs > 0, so the spring force is upward, while if the spring is compressed then ∆L < 0 and Fs < 0, so the spring force is downward. • A damping force Fd = −cy′ that resists the motion with a force proportional to the velocity of the object. It may be due to air resistance or friction in the spring. However, a convenient way to visualize a damping force is to assume that the object is rigidly attached to a piston with negligible mass immersed in a cylinder (called a dashpot) filled with a viscous liquid (Figure 6.1.2). As the piston moves, the liquid exerts a damping force. We say that the motion is undamped if c = 0, or damped if c > 0. • An external force F , other than the force due to gravity, that may vary with t, but is independent of displacement and velocity. We say that the motion is free if F ≡0, or forced if F ̸≡0. From Newton’s second law of motion, my′′ = −mg + Fd + Fs + F = −mg −cy′ + Fs + F. (6.1.1) | Elementary Differential Equations with Boundary Value Problems_Page_278_Chunk2682 |
Section 6.1 Spring Problems I 269 y (a) 0 (b) L ∆ L Figure 6.1.3 (a) Natural length of spring (b) Spring stretched by mass We must now relate Fs to y. In the absence of external forces the object stretches the spring by an amount ∆l to assume its equilibrium position (Figure 6.1.3). Since the sum of the forces acting on the object is then zero, Hooke’s Law implies that mg = k∆l. If the object is displaced y units from its equilibrium position, the total change in the length of the spring is ∆L = ∆l −y, so Hooke’s law implies that Fs = k∆L = k∆l −ky. Substituting this into (6.1.1) yields my′′ = −mg −cy′ + k∆L −ky + F. Since mg = k∆l this can be written as my′′ + cy′ + ky = F. (6.1.2) We call this the equation of motion. Simple Harmonic Motion Throughout the rest of this section we’ll consider spring–mass systems without damping; that is, c = 0. We’ll consider systems with damping in the next section. We first consider the case where the motion is also free; that is, F =0. We begin with an example. Example 6.1.1 An object stretches a spring 6 inches in equilibrium. (a) Set up the equation of motion and find its general solution. (b) Find the displacement of the object for t > 0 if it’s initially displaced 18 inches above equilibrium and given a downward velocity of 3 ft/s. | Elementary Differential Equations with Boundary Value Problems_Page_279_Chunk2683 |
270 Chapter 6 Applications of Linear Second Order Equations SOLUTION(a) Setting c = 0 and F = 0 in (6.1.2) yields the equation of motion my′′ + ky = 0, which we rewrite as y′′ + k my = 0. (6.1.3) Although we would need the weight of the object to obtain k from the equation mg = k∆l we can obtain k/m from ∆l alone; thus, k/m = g/∆l. Consistent with the units used in the problem statement, we take g = 32 ft/s2. Although ∆l is stated in inches, we must convert it to feet to be consistent with this choice of g; that is, ∆l = 1/2 ft. Therefore k m = 32 1/2 = 64 and (6.1.3) becomes y′′ + 64y = 0. (6.1.4) The characteristic equation of (6.1.4) is r2 + 64 = 0, which has the zeros r = ±8i. Therefore the general solution of (6.1.4) is y = c1 cos 8t + c2 sin 8t. (6.1.5) SOLUTION(b) The initial upward displacement of 18 inches is positive and must be expressed in feet. The initial downward velocity is negative; thus, y(0) = 3 2 and y′(0) = −3. Differentiating (6.1.5) yields y′ = −8c1 sin 8t + 8c2 cos 8t. (6.1.6) Setting t = 0 in (6.1.5) and (6.1.6) and imposing the initial conditions shows that c1 = 3/2 and c2 = −3/8. Therefore y = 3 2 cos 8t −3 8 sin8t, where y is in feet (Figure 6.1.4). We’ll now consider the equation my′′ + ky = 0 where m and k are arbitrary positive numbers. Dividing through by m and defining ω0 = p k/m yields y′′ + ω2 0y = 0. The general solution of this equation is y = c1 cos ω0t + c2 sin ω0t. (6.1.7) We can rewrite this in a more useful form by defining R = q c2 1 + c2 2, (6.1.8) | Elementary Differential Equations with Boundary Value Problems_Page_280_Chunk2684 |
Section 6.1 Spring Problems I 271 0.5 1.5 2.5 1.0 2.0 3.0 0.5 1.5 1.0 2.0 −0.5 −1.5 −1.0 −2.0 x y Figure 6.1.4 y = 3 2 cos 8t −3 8 sin 8t and c1 = R cos φ and c2 = R sin φ. (6.1.9) Substituting from (6.1.9) into (6.1.7) and applying the identity cos ω0t cos φ + sin ω0t sin φ = cos(ω0t −φ) yields y = R cos(ω0t −φ). (6.1.10) From (6.1.8) and (6.1.9) we see that the R and φ can be interpreted as polar coordinates of the point with rectangular coordinates (c1, c2) (Figure 6.1.5). Given c1 and c2, we can compute R from (6.1.8). From (6.1.8) and (6.1.9), we see that φ is related to c1 and c2 by cos φ = c1 p c2 1 + c2 2 and sin φ = c2 p c2 1 + c2 2 . There are infinitely many angles φ, differing by integer multiples of 2π, that satisfy these equations. We will always choose φ so that −π ≤φ < π. The motion described by (6.1.7) or (6.1.10) is simple harmonic motion. We see from either of these equations that the motion is periodic, with period T = 2π/ω0. This is the time required for the object to complete one full cycle of oscillation (for example, to move from its highest position to its lowest position and back to its highest position). Since the highest and lowest positions of the object are y = R and y = −R, we say that R is the amplitude of the oscillation. The | Elementary Differential Equations with Boundary Value Problems_Page_281_Chunk2685 |
272 Chapter 6 Applications of Linear Second Order Equations θ c1 c2 R Figure 6.1.5 R = p c2 1 + c2 2; c1 = R cos φ; c2 = R sin φ angle φ in (6.1.10) is the phase angle. It’s measured in radians. Equation (6.1.10) is the amplitude–phase form of the displacement. If t is in seconds then ω0 is in radians per second (rad/s); it’s the frequency of the motion. It is also called the natural frequency of the spring–mass system without damping. Example 6.1.2 We found the displacement of the object in Example 6.1.1 to be y = 3 2 cos 8t −3 8 sin8t. Find the frequency, period, amplitude, and phase angle of the motion. Solution The frequency is ω0 = 8 rad/s, and the period is T = 2π/ω0 = π/4 s. Since c1 = 3/2 and c2 = −3/8, the amplitude is R = q c2 1 + c2 2 = s3 2 2 + 3 8 2 = 3 8 √ 17. The phase angle is determined by cos φ = 3 2 3 8 √ 17 = 4 √ 17 (6.1.11) and sin φ = −3 8 3 8 √ 17 = −1 √ 17. (6.1.12) Using a calculator, we see from (6.1.11) that φ ≈±.245 rad. | Elementary Differential Equations with Boundary Value Problems_Page_282_Chunk2686 |
Section 6.1 Spring Problems I 273 Since sin φ < 0 (see (6.1.12)), the minus sign applies here; that is, φ ≈−.245 rad. Example 6.1.3 The natural length of a spring is 1 m. An object is attached to it and the length of the spring increases to 102 cm when the object is in equilibrium. Then the object is initially displaced downward 1 cm and given an upward velocity of 14 cm/s. Find the displacement for t > 0. Also, find the natural frequency, period, amplitude, and phase angle of the resulting motion. Express the answers in cgs units. Solution In cgs units g = 980 cm/s2. Since ∆l = 2 cm, ω2 0 = g/∆l = 490. Therefore y′′ + 490y = 0, y(0) = −1, y′(0) = 14. The general solution of the differential equation is y = c1 cos 7 √ 10t + c2 sin 7 √ 10t, so y′ = 7 √ 10 −c1 sin 7 √ 10t + c2 cos 7 √ 10t . Substituting the initial conditions into the last two equations yields c1 = −1 and c2 = 2/ √ 10. Hence, y = −cos 7 √ 10t + 2 √ 10 sin 7 √ 10t. The frequency is 7 √ 10 rad/s, and the period is T = 2π/(7 √ 10) s. The amplitude is R = q c2 1 + c2 2 = s (−1)2 + 2 √ 10 2 = r 7 5 cm. The phase angle is determined by cos φ = c1 R = − r 5 7 and sin φ = c2 R = r 2 7. Therefore φ is in the second quadrant and φ = cos−1 − r 5 7 ! ≈2.58 rad. Undamped Forced Oscillation In many mechanical problems a device is subjected to periodic external forces. For example, soldiers marching in cadence on a bridge cause periodic disturbances in the bridge, and the engines of a propeller driven aircraft cause periodic disturbances in its wings. In the absence of sufficient damping forces, such disturbances – even if small in magnitude – can cause structural breakdown if they are at certain critical frequencies. To illustrate, this we’ll consider the motion of an object in a spring–mass system without damping, subject to an external force F (t) = F0 cos ωt | Elementary Differential Equations with Boundary Value Problems_Page_283_Chunk2687 |
274 Chapter 6 Applications of Linear Second Order Equations where F0 is a constant. In this case the equation of motion (6.1.2) is my′′ + ky = F0 cos ωt, which we rewrite as y′′ + ω2 0y = F0 m cos ωt (6.1.13) with ω0 = p k/m. We’ll see from the next two examples that the solutions of (6.1.13) with ω ̸= ω0 behave very differently from the solutions with ω = ω0. Example 6.1.4 Solve the initial value problem y′′ + ω2 0y = F0 m cos ωt, y(0) = 0, y′(0) = 0, (6.1.14) given that ω ̸= ω0. Solution We first obtain a particular solution of (6.1.13) by the method of undetermined coefficients. Since ω ̸= ω0, cos ωt isn’t a solution of the complementary equation y′′ + ω2 0y = 0. Therefore (6.1.13) has a particular solution of the form yp = A cos ωt + B sin ωt. Since y′′ p = −ω2(A cos ωt + B sin ωt), y′′ p + ω2 0yp = F0 m cos ωt if and only if (ω2 0 −ω2) (A cos ωt + B sin ωt) = F0 m cos ωt. This holds if and only if A = F0 m(ω2 0 −ω2) and B = 0, so yp = F0 m(ω2 0 −ω2) cos ωt. The general solution of (6.1.13) is y = F0 m(ω2 0 −ω2) cos ωt + c1 cos ω0t + c2 sin ω0t, (6.1.15) so y′ = −ωF0 m(ω2 0 −ω2) sin ωt + ω0(−c1 sin ω0t + c2 cos ω0t). The initial conditions y(0) = 0 and y′(0) = 0 in (6.1.14) imply that c1 = − F0 m(ω2 0 −ω2) and c2 = 0. | Elementary Differential Equations with Boundary Value Problems_Page_284_Chunk2688 |
Section 6.1 Spring Problems I 275 Substituting these into (6.1.15) yields y = F0 m(ω2 0 −ω2)(cos ωt −cos ω0t). (6.1.16) It is revealing to write this in a different form. We start with the trigonometric identities cos(α −β) = cos α cos β + sin α sin β cos(α + β) = cos α cos β −sin α sin β. Subtracting the second identity from the first yields cos(α −β) −cos(α + β) = 2 sinα sin β (6.1.17) Now let α −β = ωt and α + β = ω0t, (6.1.18) so that α = (ω0 + ω)t 2 and β = (ω0 −ω)t 2 . (6.1.19) Substituting (6.1.18) and (6.1.19) into (6.1.17) yields cos ωt −cos ω0t = 2 sin (ω0 −ω)t 2 sin (ω0 + ω)t 2 , and substituting this into (6.1.16) yields y = R(t) sin (ω0 + ω)t 2 , (6.1.20) where R(t) = 2F0 m(ω2 0 −ω2) sin (ω0 −ω)t 2 . (6.1.21) From (6.1.20) we can regard y as a sinusoidal variation with frequency (ω0 +ω)/2 and variable ampli- tude |R(t)|. In Figure 6.1.6 the dashed curve above the t axis is y = |R(t)|, the dashed curve below the t axis is y = −|R(t)|, and the displacement y appears as an oscillation bounded by them. The oscillation of y for t on an interval between successive zeros of R(t) is called a beat. You can see from (6.1.20) and (6.1.21) that |y(t)| ≤ 2|F0| m|ω2 0 −ω2|; moreover, if ω + ω0 is sufficiently large compared with ω −ω0, then |y| assumes values close to (perhaps equal to) this upper bound during each beat. However, the oscillation remains bounded for all t. (This assumes that the spring can withstand deflections of this size and continue to obey Hooke’s law.) The next example shows that this isn’t so if ω = ω0. Example 6.1.5 Find the general solution of y′′ + ω2 0y = F0 m cos ω0t. (6.1.22) | Elementary Differential Equations with Boundary Value Problems_Page_285_Chunk2689 |
276 Chapter 6 Applications of Linear Second Order Equations t y Figure 6.1.6 Undamped oscillation with beats Solution We first obtain a particular solution yp of (6.1.22). Since cos ω0t is a solution of the comple- mentary equation, the form for yp is yp = t(A cos ω0t + B sinω0t). (6.1.23) Then y′ p = A cos ω0t + B sin ω0t + ω0t(−A sin ω0t + B cos ω0t) and y′′ p = 2ω0(−A sin ω0t + B cos ω0t) −ω2 0t(A cos ω0t + B sin ω0t). (6.1.24) From (6.1.23) and (6.1.24), we see that yp satisfies (6.1.22) if −2Aω0 sin ω0t + 2Bω0 cos ω0t = F0 m cos ω0t; that is, if A = 0 and B = F0 2mω0 . Therefore yp = F0t 2mω0 sin ω0t is a particular solution of (6.1.22). The general solution of (6.1.22) is y = F0t 2mω0 sin ω0t + c1 cos ω0t + c2 sin ω0t. | Elementary Differential Equations with Boundary Value Problems_Page_286_Chunk2690 |
Section 6.1 Spring Problems I 277 t y y = F0 t / 2mω0 y = − F0 t / 2mω0 Figure 6.1.7 Unbounded displacement due to resonance The graph of yp is shown in Figure 6.1.7, where it can be seen that yp oscillates between the dashed lines y = F0t 2mω0 and y = −F0t 2mω0 with increasing amplitude that approaches ∞as t →∞. Of course, this means that the spring must eventually fail to obey Hooke’s law or break. This phenomenon of unbounded displacements of a spring–mass system in response to a periodic forcing function at its natural frequency is called resonance. More complicated mechanical structures can also exhibit resonance–like phenomena. For example, rhythmic oscillations of a suspension bridge by wind forces or of an airplane wing by periodic vibrations of reciprocating engines can cause damage or even failure if the frequencies of the disturbances are close to critical frequencies determined by the parameters of the mechanical system in question. 6.1 Exercises In the following exercises assume that there’s no damping. 1. C/G An object stretches a spring 4 inches in equilibrium. Find and graph its displacement for t > 0 if it’s initially displaced 36 inches above equilibrium and given a downward velocity of 2 ft/s. 2. An object stretches a string 1.2 inches in equilibrium. Find its displacement for t > 0 if it’s initially displaced 3 inches below equilibrium and given a downward velocity of 2 ft/s. 3. A spring with natural length .5 m has length 50.5 cm with a mass of 2 gm suspended from it. The mass is initially displaced 1.5 cm below equilibrium and released with zero velocity. Find its displacement for t > 0. | Elementary Differential Equations with Boundary Value Problems_Page_287_Chunk2691 |
278 Chapter 6 Applications of Linear Second Order Equations 4. An object stretches a spring 6 inches in equilibrium. Find its displacement for t > 0 if it’s initially displaced 3 inches above equilibrium and given a downward velocity of 6 inches/s. Find the frequency, period, amplitude and phase angle of the motion. 5. C/G An object stretches a spring 5 cm in equilibrium. It is initially displaced 10 cm above equilibrium and given an upward velocity of .25 m/s. Find and graph its displacement for t > 0. Find the frequency, period, amplitude, and phase angle of the motion. 6. A 10 kg mass stretches a spring 70 cm in equilibrium. Suppose a 2 kg mass is attached to the spring, initially displaced 25 cm below equilibrium, and given an upward velocity of 2 m/s. Find its displacement for t > 0. Find the frequency, period, amplitude, and phase angle of the motion. 7. A weight stretches a spring 1.5 inches in equilibrium. The weight is initially displaced 8 inches above equilibrium and given a downward velocity of 4 ft/s. Find its displacement for t > 0. 8. A weight stretches a spring 6 inches in equilibrium. The weight is initially displaced 6 inches above equilibrium and given a downward velocity of 3 ft/s. Find its displacement for t > 0. 9. A spring–mass system has natural frequency 7 √ 10 rad/s. The natural length of the spring is .7 m. What is the length of the spring when the mass is in equilibrium? 10. A 64 lb weight is attached to a spring with constant k = 8 lb/ft and subjected to an external force F (t) = 2 sin t. The weight is initially displaced 6 inches above equilibrium and given an upward velocity of 2 ft/s. Find its displacement for t > 0. 11. A unit mass hangs in equilibrium from a spring with constant k = 1/16. Starting at t = 0, a force F (t) = 3 sint is applied to the mass. Find its displacement for t > 0. 12. C/G A 4 lb weight stretches a spring 1 ft in equilibrium. An external force F (t) = .25 sin8t lb is applied to the weight, which is initially displaced 4 inches above equilibrium and given a downward velocity of 1 ft/s. Find and graph its displacement for t > 0. 13. A 2 lb weight stretches a spring 6 inches in equilibrium. An external force F (t) = sin 8t lb is ap- plied to the weight, which is released from rest 2 inches below equilibrium. Find its displacement for t > 0. 14. A 10 gm mass suspended on a spring moves in simple harmonic motion with period 4 s. Find the period of the simple harmonic motion of a 20 gm mass suspended from the same spring. 15. A 6 lb weight stretches a spring 6 inches in equilibrium. Suppose an external force F (t) = 3 16 sin ωt + 3 8 cos ωt lb is applied to the weight. For what value of ω will the displacement be unbounded? Find the displacement if ω has this value. Assume that the motion starts from equilibrium with zero initial velocity. 16. C/G A 6 lb weight stretches a spring 4 inches in equilibrium. Suppose an external force F (t) = 4 sinωt −6 cos ωt lb is applied to the weight. For what value of ω will the displacement be unbounded? Find and graph the displacement if ω has this value. Assume that the motion starts from equilibrium with zero initial velocity. 17. A mass of one kg is attached to a spring with constant k = 4 N/m. An external force F (t) = −cos ωt −2 sin ωt n is applied to the mass. Find the displacement y for t > 0 if ω equals the natural frequency of the spring–mass system. Assume that the mass is initially displaced 3 m above equilibrium and given an upward velocity of 450 cm/s. 18. An object is in simple harmonic motion with frequency | Elementary Differential Equations with Boundary Value Problems_Page_288_Chunk2692 |
ω0, with y(0) = y0 and y′(0) = v0. Find its displacement for t > 0. Also, find the amplitude of the oscillation and give formulas for the sine and cosine of the initial phase angle. | Elementary Differential Equations with Boundary Value Problems_Page_288_Chunk2693 |
Section 6.2 Spring Problems II 279 19. Two objects suspended from identical springs are set into motion. The period of one object is twice the period of the other. How are the weights of the two objects related? 20. Two objects suspended from identical springs are set into motion. The weight of one object is twice the weight of the other. How are the periods of the resulting motions related? 21. Two identical objects suspended from different springs are set into motion. The period of one motion is 3 times the period of the other. How are the two spring constants related? 6.2 SPRING PROBLEMS II Free Vibrations With Damping In this section we consider the motion of an object in a spring–mass system with damping. We start with unforced motion, so the equation of motion is my′′ + cy′ + ky = 0. (6.2.1) Now suppose the object is displaced from equilibrium and given an initial velocity. Intuition suggests that if the damping force is sufficiently weak the resulting motion will be oscillatory, as in the undamped case considered in the previous section, while if it’s sufficiently strong the object may just move slowly toward the equilibrium position without ever reaching it. We’ll now confirm these intuitive ideas mathematically. The characteristic equation of (6.2.1) is mr2 + cr + k = 0. The roots of this equation are r1 = −c − √ c2 −4mk 2m and r2 = −c + √ c2 −4mk 2m . (6.2.2) In Section 5.2 we saw that the form of the solution of (6.2.1) depends upon whether c2 −4mk is positive, negative, or zero. We’ll now consider these three cases. Underdamped Motion We say the motion is underdamped if c < √ 4mk. In this case r1 and r2 in (6.2.2) are complex conjugates, which we write as r1 = −c 2m −iω1 and r2 = −c 2m + iω1, where ω1 = √ 4mk −c2 2m . The general solution of (6.2.1) in this case is y = e−ct/2m(c1 cos ω1t + c2 sin ω1t). By the method used in Section 6.1 to derive the amplitude–phase form of the displacement of an object in simple harmonic motion, we can rewrite this equation as y = Re−ct/2m cos(ω1t −φ), (6.2.3) | Elementary Differential Equations with Boundary Value Problems_Page_289_Chunk2694 |
280 Chapter 6 Applications of Linear Second Order Equations x y y = Re−ct / 2m y = −− Re−ct / 2m Figure 6.2.1 Underdamped motion where R = q c2 1 + c2 2, R cos φ = c1, and R sin φ = c2. The factor Re−ct/2m in (6.2.3) is called the time–varying amplitude of the motion, the quantity ω1 is called the frequency, and T = 2π/ω1 (which is the period of the cosine function in (6.2.3) is called the quasi–period. A typical graph of (6.2.3) is shown in Figure 6.2.1. As illustrated in that figure, the graph of y oscillates between the dashed exponential curves y = ±Re−ct/2m. | Elementary Differential Equations with Boundary Value Problems_Page_290_Chunk2695 |
Section 6.2 Spring Problems II 281 Overdamped Motion We say the motion is overdamped if c > √ 4mk. In this case the zeros r1 and r2 of the characteristic polynomial are real, with r1 < r2 < 0 (see (6.2.2)), and the general solution of (6.2.1) is y = c1er1t + c2er2t. Again limt→∞y(t) = 0 as in the underdamped case, but the motion isn’t oscillatory, since y can’t equal zero for more than one value of t unless c1 = c2 = 0. (Exercise 23.) Critically Damped Motion We say the motion is critically damped if c = √ 4mk. In this case r1 = r2 = −c/2m and the general solution of (6.2.1) is y = e−ct/2m(c1 + c2t). Again limt→∞y(t) = 0 and the motion is nonoscillatory, since y can’t equal zero for more than one value of t unless c1 = c2 = 0. (Exercise 22). Example 6.2.1 Suppose a 64 lb weight stretches a spring 6 inches in equilibrium and a dashpot provides a damping force of c lb for each ft/sec of velocity. (a) Write the equation of motion of the object and determine the value of c for which the motion is critically damped. (b) Find the displacement y for t > 0 if the motion is critically damped and the initial conditions are y(0) = 1 and y′(0) = 20. (c) Find the displacement y for t > 0 if the motion is critically damped and the initial conditions are y(0) = 1 and y′(0) = −20. SOLUTION(a) Here m = 2 slugs and k = 64/.5 = 128 lb/ft. Therefore the equation of motion (6.2.1) is 2y′′ + cy′ + 128y = 0. (6.2.4) The characteristic equation is 2r2 + cr + 128 = 0, which has roots r = −c ± √ c2 −8 · 128 4 . Therefore the damping is critical if c = √ 8 · 128 = 32 lb–sec/ft. SOLUTION(b) Setting c = 32 in (6.2.4) and cancelling the common factor 2 yields y′′ + 16y + 64y = 0. The characteristic equation is r2 + 16r + 64y = (r + 8)2 = 0. Hence, the general solution is y = e−8t(c1 + c2t). (6.2.5) | Elementary Differential Equations with Boundary Value Problems_Page_291_Chunk2696 |
282 Chapter 6 Applications of Linear Second Order Equations 0.2 0.4 0.6 0.8 1.0 0.5 1.5 1.0 2.0 −0.5 (a) (b) x y Figure 6.2.2 (a) y = e−8t(1 + 28t) (b) y = e−8t(1 −12t) Differentiating this yields y′ = −8y + c2e−8t. (6.2.6) Imposing the initial conditions y(0) = 1 and y′(0) = 20 in the last two equations shows that 1 = c1 and 20 = −8 + c2. Hence, the solution of the initial value problem is y = e−8t(1 + 28t). Therefore the object approaches equilibrium from above as t →∞. There’s no oscillation. SOLUTION(c) Imposing the initial conditions y(0) = 1 and y′(0) = −20 in (6.2.5) and (6.2.6) yields 1 = c1 and −20 = −8 + c2. Hence, the solution of this initial value problem is y = e−8t(1 −12t). Therefore the object moves downward through equilibrium just once, and then approaches equilibrium from below as t →∞. Again, there’s no oscillation. The solutions of these two initial value problems are graphed in Figure 6.2.2. Example 6.2.2 Find the displacement of the object in Example 6.2.1 if the damping constant is c = 4 lb–sec/ft and the initial conditions are y(0) = 1.5 ft and y′(0) = −3 ft/sec. Solution With c = 4, the equation of motion (6.2.4) becomes y′′ + 2y′ + 64y = 0 (6.2.7) after cancelling the common factor 2. The characteristic equation r2 + 2r + 64 = 0 | Elementary Differential Equations with Boundary Value Problems_Page_292_Chunk2697 |
Section 6.2 Spring Problems II 283 has complex conjugate roots r = −2 ± √4 −4 · 64 2 = −1 ± 3 √ 7i. Therefore the motion is underdamped and the general solution of (6.2.7) is y = e−t(c1 cos 3 √ 7t + c2 sin 3 √ 7t). Differentiating this yields y′ = −y + 3 √ 7e−t(−c1 sin 3 √ 7t + c2 cos 3 √ 7t). Imposing the initial conditions y(0) = 1.5 and y′(0) = −3 in the last two equations yields 1.5 = c1 and −3 = −1.5 + 3 √ 7c2. Hence, the solution of the initial value problem is y = e−t 3 2 cos 3 √ 7t − 1 2 √ 7 sin 3 √ 7t . (6.2.8) The amplitude of the function in parentheses is R = s3 2 2 + 1 2 √ 7 2 = r 9 4 + 1 4 · 7 = r 64 4 · 7 = 4 √ 7. Therefore we can rewrite (6.2.8) as y = 4 √ 7e−t cos(3 √ 7t −φ), where cos φ = 3 2R = 3 √ 7 8 and sin φ = − 1 2 √ 7R = −1 8. Therefore φ ∼= −.125 radians. Example 6.2.3 Let the damping constant in Example 1 be c = 40 lb–sec/ft. Find the displacement y for t > 0 if y(0) = 1 and y′(0) = 1. Solution With c = 40, the equation of motion (6.2.4) reduces to y′′ + 20y′ + 64y = 0 (6.2.9) after cancelling the common factor 2. The characteristic equation r2 + 20r + 64 = (r + 16)(r + 4) = 0 has the roots r1 = −4 and r2 = −16. Therefore the general solution of (6.2.9) is y = c1e−4t + c2e−16t. (6.2.10) Differentiating this yields y′ = −4e−4t −16c2e−16t. | Elementary Differential Equations with Boundary Value Problems_Page_293_Chunk2698 |
284 Chapter 6 Applications of Linear Second Order Equations 0.2 0.4 0.6 0.8 1.0 1.2 0.2 0.4 0.6 0.8 1.0 x y Figure 6.2.3 y = 17 12e−4t −5 12e−16t The last two equations and the initial conditions y(0) = 1 and y′(0) = 1 imply that c1 + c2 = 1 −4c1 − 16c2 = 1. The solution of this system is c1 = 17/12, c2 = −5/12. Substituting these into (6.2.10) yields y = 17 12e−4t −5 12e−16t as the solution of the given initial value problem (Figure 6.2.3). Forced Vibrations With Damping Now we consider the motion of an object in a spring-mass system with damping, under the influence of a periodic forcing function F (t) = F0 cos ωt, so that the equation of motion is my′′ + cy′ + ky = F0 cos ωt. (6.2.11) In Section 6.1 we considered this equation with c = 0 and found that the resulting displacement y assumed arbitrarily large values in the case of resonance (that is, when ω = ω0 = p k/m). Here we’ll see that in the presence of damping the displacement remains bounded for all t, and the initial conditions have little effect on the motion as t →∞. In fact, we’ll see that for large t the displacement is closely approximated by a function of the form y = R cos(ωt −φ), (6.2.12) where the amplitude R depends upon m, c, k, F0, and ω. We’re interested in the following question: | Elementary Differential Equations with Boundary Value Problems_Page_294_Chunk2699 |
Section 6.2 Spring Problems II 285 QUESTION:Assuming that m, c, k, and F0 are held constant, what value of ω produces the largest amplitude R in (6.2.12), and what is this largest amplitude? To answer this question, we must solve (6.2.11) and determine R in terms of F0, ω0, ω, and c. We can obtain a particular solution of (6.2.11) by the method of undetermined coefficients. Since cos ωt does not satisfy the complementary equation my′′ + cy′ + ky = 0, we can obtain a particular solution of (6.2.11) in the form yp = A cos ωt + B sin ωt. (6.2.13) Differentiating this yields y′ p = ω(−A sin ωt + B cos ωt) and y′′ p = −ω2(A cos ωt + B sin ωt). From the last three equations, my′′ p + cy′ p + kyp = (−mω2A + cωB + kA) cos ωt + (−mω2B −cωA + kB) sin ωt, so yp satisfies (6.2.11) if (k −mω2)A + cωB = F0 −cωA + (k −mω2)B = 0. Solving for A and B and substituting the results into (6.2.13) yields yp = F0 (k −mω2)2 + c2ω2 (k −mω2) cos ωt + cω sin ωt , which can be written in amplitude–phase form as yp = F0 p (k −mω2)2 + c2ω2 cos(ωt −φ), (6.2.14) where cos φ = k −mω2 p (k −mω2)2 + c2ω2 and sin φ = cω p (k −mω2)2 + c2ω2 . (6.2.15) To compare this with the undamped forced vibration that we considered in Section 6.1 it’s useful to write k −mω2 = m k m −ω2 = m(ω2 0 −ω2), (6.2.16) where ω0 = p k/m is the natural angular frequency of the undamped simple harmonic motion of an object with mass m on a spring with constant k. Substituting (6.2.16) into (6.2.14) yields yp = F0 p m2(ω2 0 −ω2)2 + c2ω2 cos(ωt −φ). (6.2.17) The solution of an initial value problem my′′ + cy′ + ky = F0 cos ωt, y(0) = y0, y′(0) = v0, | Elementary Differential Equations with Boundary Value Problems_Page_295_Chunk2700 |
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