text stringlengths 1 7.76k | source stringlengths 17 81 |
|---|---|
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 ... | 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).... | 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 solvi... | 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|... | 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 fam... | 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 .... | 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) t... | 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... | 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 ... | 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 =... | 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. Therefor... | 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′... | 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 initi... | 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 = c1e... | 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 ... | 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 th... | 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... | 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 a... | 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 (−∞, ∞). ... | 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 anothe... | 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 s... | 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), ... | 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 combi... | 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 ... | 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, ... | 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... | 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 po... | 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... | 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... | 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′′ = ... | 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 ... | 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. The... | 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. ... | 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 eq... | 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 ... | 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 th... | 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 +... | 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 continuo... | 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 i... | 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... | 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 Exercis... | 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... | 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 ... | 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... | 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). T... | 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 ... | 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, ... | 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 obt... | 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 und... | 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... | 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), ... | 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 ... | 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 c... | 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... | 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 polynomialsmu... | 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 thi... | 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) sho... | 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 ... | 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′′ ... | 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 ω... | 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 + · · ·... | 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) ... | 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 ) + P... | 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... | 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 ... | 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 ord... | 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 ... | 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... | 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 differenti... | 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 p... | 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 (... | 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 complement... | 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 ... | 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 ... | 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 + ... | 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−... | 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 = 4x... | 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′ −... | 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. F... | 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 fo... | 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... | 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 o... | 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... | 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... | 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... | 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 displace... | 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 d... | 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 identi... | 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 ... | 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 co... | 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.... | 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 ... | 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... | 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 ... | 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 =... | 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 cond... | 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. Substi... | 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... | Elementary Differential Equations with Boundary Value Problems_Page_295_Chunk2700 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.