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Section 13.2 Sturm-Liouville Problems 689 In this case the general solution of the differential equation in (13.2.2) is y = e−3x/2(c1 cos ωx + c2 sinωx). The boundary condition y(0) = 0 requires that c1 = 0, so y = c2e−3x/2 sin ωx, which holds with c2 ̸= 0 if and only if ω = nπ, where n is an integer. We may assume that n is a positive integer. (Why?). From (13.2.3), the eigenvalues are λn = (1 + 4n2π2)/4, with associated eigenfunctions yn = e−3x/2 sin nπx, n = 1, 2, 3, . . .. Example 13.2.2 Solve the eigenvalue problem x2y′′ + xy′ + λy = 0, y(1) = 0, y(2) = 0. (13.2.4) Solution If λ = 0, the differential equation in (13.2.4) reduces to x(xy′)′ = 0, so xy′ = c1, y′ = c1 x , and y = c1 ln x + c2. The boundary condition y(1) = 0 requires that c2 = 0, so y = c1 ln x. The boundary condition y(2) = 0 requires that c1 ln2 = 0, so c1 = 0. Therefore zero isn’t an eigenvalue of (13.2.4). If λ < 0, we write λ = −k2 with k > 0, so (13.2.4) becomes x2y′′ + xy′ −k2y = 0, an Euler equation (Section 7.4) with indicial equation r2 −k2 = (r −k)(r + k) = 0. Therefore y = c1xk + c2x−k. The boundary conditions require that c1 + c2 = 0 2kc1 + 2−kc2 = 0. Since the determinant of this system is 2−k −2k ̸= 0, c1 = c2 = 0. Therefore (13.2.4) has no negative eigenvalues. If λ > 0 we write λ = k2 with k > 0. Then (13.2.4) becomes x2y′′ + xy′ + k2y = 0, an Euler equation with indicial equation r2 + k2 = (r −ik)(r + ik) = 0, so y = c1 cos(k lnx) + c2 sin(k ln x). The boundary condition y(1) = 0 requires that c1 = 0. Therefore y = c2 sin(k ln x). This holds with c2 ̸= 0 if and only if k = nπ/ ln2, where n is a positive integer. Hence, the eigenvalues of (13.2.4) are λn = (nπ/ ln2)2, with associated eigenfunctions yn = sin nπ ln2 lnx , n = 1, 2, 3, . . .. For theoretical purposes, it’s useful to rewrite the differential equation in (13.2.1) in a different form, provided by the next theorem. | Elementary Differential Equations with Boundary Value Problems_Page_699_Chunk3101 |
690 Chapter 13 Boundary Value Problems for Second Order Ordinary Differential Equations Theorem 13.2.1 If P0, P1, P2, and R are continuous and P0 and R are positive on a closed interval [a, b], then the equation P0(x)y′′ + P1(x)y′ + P2(x)y + λR(x)y = 0 (13.2.5) can be rewritten as (p(x)y′)′ + q(x)y + λr(x)y = 0, (13.2.6) where p, p′, q and r are continuous and p and r are positive on [a, b]. Proof We begin by rewriting (13.2.5) as y′′ + u(x)y′ + v(x)y + λR1(x)y = 0, (13.2.7) with u = P1/P0, v = P2/P0, and R1 = R/P0. (Note that R1 is positive on [a, b].) Now let p(x) = eU(x), where U is any antiderivative of u. Then p is positive on [a, b] and, since U ′ = u, p′(x) = p(x)u(x) (13.2.8) is continuous on [a, b]. Multiplying (13.2.7) by p(x) yields p(x)y′′ + p(x)u(x)y′ + p(x)v(x)y + λp(x)R1(x)y = 0. (13.2.9) Since p is positive on [a, b], this equation has the same solutions as (13.2.5). From (13.2.8), (p(x)y′)′ = p(x)y′′ + p′(x)y′ = p(x)y′′ + p(x)u(x)y′, so (13.2.9) can be rewritten as in (13.2.6), with q(x) = p(x)v(x) and r(x) = p(x)R1(x). This completes the proof. It is to be understood throughout the rest of this section that p, q, and r have the properties stated in Theorem 13.2.1. Moreover, whenever we write Ly in a general statement, we mean Ly = (p(x)y′)′ + q(x)y. The differential equation (13.2.6) is called a Sturm–Liouville equation, and the eigenvalue problem (p(x)y′)′ + q(x)y + λr(x)y = 0, B1(y) = 0, B2(y) = 0, (13.2.10) which is equivalent to (13.2.1), is called a Sturm-Liouville problem. Example 13.2.3 Rewrite the eigenvalue problem y′′ + 3y′ + (2 + λ)y = 0, y(0) = 0, y(1) = 0 (13.2.11) of Example 13.2.1 as a Sturm-Liouville problem. Solution Comparing (13.2.11) to (13.2.7) shows that u(x) = 3, so we take U(x) = 3x and p(x) = e3x. Multiplying the differential equation in (13.2.11) by e3x yields e3x(y′′ + 3y′) + 2e3xy + λe3xy = 0. Since e3x(y′′ + 3y′) = (e3xy′)′, (13.2.11) is equivalent to the Sturm–Liouville problem (e3xy′)′ + 2e3xy + λe3xy = 0, y(0) = 0, y(1) = 0. (13.2.12) | Elementary Differential Equations with Boundary Value Problems_Page_700_Chunk3102 |
Section 13.2 Sturm-Liouville Problems 691 Example 13.2.4 Rewrite the eigenvalue problem x2y′′ + xy′ + λy = 0, y(1) = 0, y(2) = 0 (13.2.13) of Example 13.2.2 as a Sturm-Liouville problem. Solution Dividing the differential equation in (13.2.13) by x2 yields y′′ + 1 xy′ + λ x2 y = 0. Comparing this to (13.2.7) shows that u(x) = 1/x, so we take U(x) = ln x and p(x) = eln x = x. Multiplying the differntial equation by x yields xy′′ + y′ + λ x y = 0. Since xy′′ + y′ = (xy′)′, (13.2.13) is equivalent to the Sturm–Liouville problem (xy′)′ + λ xy = 0, y(1) = 0, y(2) = 0. (13.2.14) Problems 1–4 of Section 11.1 are Sturm–Liouville problems. (Problem 5 isn’t , although some authors use a definition of Sturm-Liouville problem that does include it.) We were able to find the eigenvalues of Problems 1-4 explicitly because in each problem the coefficients in the boundary conditions satisfy αβ = 0 and ρδ = 0; that is, each boundary condition involves either y or y′, but not both. If this isn’t true then the eigenvalues can’t in general be expressed exactly by simple formulas; rather, approximate values must be obtained by numerical solution of equations derived by requiring the determinants of certain 2 × 2 systems of homogeneous equations to be zero. To apply the numerical methods effectively, graphical methods must be used to determine approximate locations of the zeros of these determinants. Then the zeros can be computed accurately by numerical methods. Example 13.2.5 Solve the Sturm–Liouville problem y′′ + λy = 0, y(0) + y′(0) = 0, y(1) + 3y′(1) = 0. (13.2.15) Solution If λ = 0, the differential equation in (13.2.15) reduces to y′′ = 0, with general solution y = c1 + c2x. The boundary conditions require that c1 + c2 = 0 c1 + 4c2 = 0, so c1 = c2 = 0. Therefore zero isn’t an eigenvalue of (13.2.15). If λ < 0, we write λ = −k2 where k > 0, and the differential equation in (13.2.15) becomes y′′ − k2y = 0, with general solution y = c1 cosh kx + c2 sinh kx, (13.2.16) so y′ = k(c1 sinh kx + c2 cosh kx). | Elementary Differential Equations with Boundary Value Problems_Page_701_Chunk3103 |
692 Chapter 13 Boundary Value Problems for Second Order Ordinary Differential Equations 1 2 3 4 1 2 3 4 k u Figure 13.2.1 u = tanh k and u = −2k/(1 −3k2) The boundary conditions require that c1 + kc2 = 0 (cosh k + 3k sinh k)c1 + (sinh k + 3k cosh k)c2 = 0. (13.2.17) The determinant of this system is DN(k) = 1 k cosh k + 3k sinhk sinh k + 3k cosh k = (1 −3k2) sinh k + 2k cosh k. Therefore the system (13.2.17) has a nontrivial solution if and only if DN(k) = 0 or, equivalently, tanh k = − 2k 1 −3k2 . (13.2.18) The graph of the right side (Figure 13.2.1) has a vertical asymptote at k = 1/ √ 3. Since the two sides have different signs if k < 1/ √ 3, this equation has no solution in (0, 1/ √ 3). Figure 13.2.1 shows the graphs of the two sides of (13.2.18) on an interval to the right of the vertical asymptote, which is indicated by the dashed line. You can see that the two curves intersect near k0 = 1.2, Given this estmate, you can use Newton’s to compute k0 more accurately. We computed k0 ≈1.1219395. Therefore −k2 0 ≈−1.2587483 is an eigenvalue of (13.2.15). From (13.2.16) and the first equation in (13.2.17), y0 = k0 cosh k0x −sinh k0x. If λ > 0 we write λ = k2 where k > 0, and differential equation in (13.2.15) becomes y′′ + k2y = 0, with general solution y = cos kx + c2 sin kx, (13.2.19) | Elementary Differential Equations with Boundary Value Problems_Page_702_Chunk3104 |
Section 13.2 Sturm-Liouville Problems 693 so y′ = k(−c1 sinkx + c2 cos kx). The boundary conditions require that c1 + kc2 = 0 (cos k −3k sin k)c1 + (sin k + 3k cos k)c2 = 0. (13.2.20) The determinant of this system is DP (k) = 1 k cos k −3k sin k sin k + 3k cos k = (1 + 3k2) sin k + 2k cos k. The system (13.2.20) has a nontrivial solution if and only if DP (k) = 0 or, equivalently, tan k = − 2k 1 + 3k2 . Figure 13.2.2 shows the graphs of the two sides of this equation. You can see from the figure that the graphs intersect at infinitely many points kn ≈nπ (n = 1, 2, 3,...), where the error in this approximation approaches zero as n →∞. Given this estimate, you can use Newton’s method to compute kn more accurately. We computed k1 ≈ 2.9256856, k2 ≈ 6.1765914, k3 ≈ 9.3538959, k4 ≈ 12.5132570. The estimates of the corresponding eigenvalues λn = k2 n are λ1 ≈ 8.5596361, λ2 ≈ 38.1502809, λ3 ≈ 87.4953676, λ4 ≈ 156.5815998. From (13.2.19) and the first equation in (13.2.20), yn = kn cos knx −sin knx is an eigenfunction associated with λn Since the differential equations in (13.2.12) and (13.2.14) are more complicated than those in (13.2.11) and (13.2.13) respectively, what is the point of Theorem 13.2.1? The point is this: to solve a specific problem, it may be better to deal with it directly, as we did in Examples 13.2.1 and 13.2.2; however, we’ll see that transforming the general eigenvalue problem (13.2.1) to the Sturm–Liouville problem (13.2.10) leads to results applicable to all eigenvalue problems of the form (13.2.1). Theorem 13.2.2 If Ly = (p(x)y′)′ + q(x)y and u and v are twice continuously functions on [a, b] that satisfy the boundary conditions B1(y) = 0 and B2(y) = 0, then Z b a [u(x)Lv(x) −v(x)Lu(x)] dx = 0. (13.2.21) | Elementary Differential Equations with Boundary Value Problems_Page_703_Chunk3105 |
694 Chapter 13 Boundary Value Problems for Second Order Ordinary Differential Equations k = π k = 2π k = 3π k = 4π k u 1 − 1 Figure 13.2.2 u = tan k and u = −2k/(1 + k) Proof Integration by parts yields Z b a [u(x)Lv(x) −v(x)Lu(x)] dx = Z b a [u(x)(p(x)v′(x))′ −v(x)(p(x)u′(x))′] dx = p(x)[u(x)v′(x) −u′(x)v(x)] b a − Z b a p(x)[u′(x)v′(x) −u′(x)v′(x)] dx. Since the last integral equals zero, Z b a [u(x)Lv(x) −v(x)Lu(x)] dx = p(x)[u(x)v′(x) −u′(x)v′(x)] b a . (13.2.22) By assumption, B1(u) = B1(v) = 0 and B2(u) = B2(v) = 0. Therefore αu(a) + βu′(a) = 0 αv(a) + βv′(a) = 0 and ρu(b) + δu′(b) = 0 ρv(b) + δv′(b) = 0. Since α2 + β2 > 0 and ρ2 + δ2 > 0, the determinants of these two systems must both be zero; that is, u(a)v′(a) −u′(a)v(a) = u(b)v′(b) −u′(b)v(b) = 0. This and (13.2.22) imply (13.2.21), which completes the proof. The next theorem shows that a Sturm–Liouville problem has no complex eigenvalues. | Elementary Differential Equations with Boundary Value Problems_Page_704_Chunk3106 |
Section 13.2 Sturm-Liouville Problems 695 Theorem 13.2.3 If λ = p + qi with q ̸= 0 then the boundary value problem Ly + λr(x)y = 0, B1(y) = 0, B2(y) = 0 has only the trivial solution. Proof For this theorem to make sense, we must consider complex-valued solutions of Ly + (p + iq)r(x, y)y = 0. (13.2.23) If y = u + iv where u and v are real-valued and twice differentiable, we define y′ = u′ + iv′ and y′′ = u′′ + iv′′. We say that y is a solution of (13.2.23) if the real and imaginary parts of the left side of (13.2.23) are both zero. Since Ly = (p(x)′y)′ + q(x)y and p, q, and r are real-valued, Ly + λr(x)y = L(u + iv) + (p + iq)r(x)(u + iv) = Lu + r(x)(pu −qv) + i[Lv + r(x)(pu + qv)], so Ly + λr(x)y = 0 if and only if Lu + r(x)(pu −qv) = 0 Lv + r(x)(qu + pv) = 0. Multiplying the first equation by v and the second by u yields vLu + r(x)(puv −qv2) = 0 uLv + r(x)(qu2 + puv) = 0. Subtracting the first equation from the second yields uLv −vLu + qr(x)(u2 + v2) = 0, so Z b a [u(x)Lv(x) −v(x)Lu(x)] dx + Z b a r(x)[u2(x) + v2(x)] dx = 0. (13.2.24) Since B1(y) = B1(u + iv) = B1(u) + iB1(v) and B2(y) = B2(u + iv) = B2(u) + iB2(v), B1(y) = 0 and B2(y) = 0 implies that B1(u) = B2(u) = B1(v) = B2(v) = 0. Therefore Theorem 13.2.2 implies that first integral in (13.2.24) equals zero, so (13.2.24) reduces to q Z b a r(x)[u2(x) + v2(x)] dx = 0. Since r is positive on [a, b] and q ̸= 0 by assumption, this implies that u ≡0 and v ≡0 on [a, b]. Therefore y ≡0 on [a, b], which completes the proof. | Elementary Differential Equations with Boundary Value Problems_Page_705_Chunk3107 |
696 Chapter 13 Boundary Value Problems for Second Order Ordinary Differential Equations Theorem 13.2.4 If λ1 and λ2 are distinct eigenvalues of the Sturm–Liouville problem Ly + λr(x)y = 0, B1(y) = 0, B2(y) = 0 (13.2.25) with associated eigenfunctions u and v respectively, then Z b a r(x)u(x)v(x) dx = 0. (13.2.26) Proof Since u and v satisfy the boundary conditions in (13.2.25), Theorem 13.2.2 implies that Z b a [u(x)Lv(x) −v(x)Lu(x)] dx = 0. Since Lu = −λ1ru and Lv = −λ2rv, this implies that (λ1 −λ2) Z b a r(x)u(x)v(x) dx = 0. Since λ1 ̸= λ2, this implies (13.2.26), which completes the proof. If u and v are any integrable functions on [a, b] and Z b a r(x)u(x)v(x) dx = 0, we say that u and v orthogonal on [a, b] with respect to r = r(x). Theorem 13.1.1 implies the next theorem. Theorem 13.2.5 If u ̸≡0 and v both satisfy Ly + λr(x)y = 0, B1(y) = 0, B2(y) = 0, then v = cu for some constant c. We’ve now proved parts of the next theorem. A complete proof is beyond the scope of this book. Theorem 13.2.6 The set of all eigenvalues of the Sturm–Liouville problem Ly + λr(x)y = 0, B1(y) = 0, B2(y) = 0 can be ordered as λ1 < λ2 < · · · < λn < · · · , and lim n→∞λn = ∞. For each n, if yn is an arbitrary λn-eigenfunction, then every λn-eigenfunction is a constant multiple of yn. If m ̸= n, ym and yn are orthogonal [a, b] with respect to r = r(x); that is, Z b a r(x)ym(x)yn(x) dx = 0. (13.2.27) You may want to verify (13.2.27) for the eigenfunctions obtained in Examples 13.2.1 and 13.2.2. In conclusion, we mention the next theorem. The proof is beyond the scope of this book. | Elementary Differential Equations with Boundary Value Problems_Page_706_Chunk3108 |
Section 13.2 Sturm-Liouville Problems 697 Theorem 13.2.7 Let λ1 < λ2 < · · · < λn < · · · be the eigenvalues of the Sturm–Liouville problem Ly + λr(x)y = 0, B1(y) = 0, B2(y) = 0, with associated eigenvectors y1, y2, ..., yn, .... Suppose f is piecewise smooth (Definition 11.2.3) on [a, b]. For each n, let cn = Z b a r(x)f(x)yn(x) dx Z b a r(x)y2 n(x) dx . Then f(x−) + f(x+) 2 = ∞ X n=1 cnyn(x) for all x in the open interval (a, b). 13.2 Exercises In Exercises 1–7 rewrite the equation in Sturm–Liouville form (with λ = 0). Assume that b, c, α, and ν are contants. 1. y′′ + by′ + cy = 0 2. x2y′′ + xy′ + (x2 −ν2)y = 0 (Bessel’s equation) 3. (1 −x2)y′′ −xy′ + α2y = 0 (Chebyshev’s equation) 4. x2y′′ + bxy′ + cy = 0 (Euler’s equation) 5. y′′ −2xy′ + 2αy = 0 (Hermite’s equation) 6. xy′′ + (1 −x)y′ + αy = 0 (Laguerre’s equation) 7. (1 −x2)y′′ −2xy′ + α(α + 1)y = 0 (Legendre’s equation) 8. In Example 13.2.4 we found that the eigenvalue problem x2y′′ + xy′ + λy = 0, y(1) = 0, y(2) = 0 (A) is equivalent to the Sturm-Liouville problem (xy′)′ + λ xy = 0, y(1) = 0, y(2) = 0. (B) Multiply the differential equation in (B) by y and integrate to show that λ Z 2 1 y2(x) x dx = Z 2 1 x(y′(x))2 dx. Conclude from this that the eigenvalues of (A) are all positive. 9. Solve the eigenvalue problem y′′ + 2y′ + y + λy = 0, y(0) = 0, y(1) = 0. | Elementary Differential Equations with Boundary Value Problems_Page_707_Chunk3109 |
698 Chapter 13 Boundary Value Problems for Second Order Ordinary Differential Equations 10. Solve the eigenvalue problem y′′ + 2y′ + y + λy = 0, y′(0) = 0, y′(1) = 0. In Exercises 11–20 : (a) Determine whether λ = 0 is an eigenvalue. If it is, find an associated eigenfunc- tion. (b) Compute the negative eigenvalues with errors not greater than 5 × 10−8. State the form of the associated eigenfunctions. (c) Compute the first four positive eigenvalues with errors not greater than 5 × 10−8. State the form of the associated eigenfunctions. 11. C y′′ + λy = 0, y(0) + 2y′(0) = 0, y(2) = 0 12. C y′′ + λy = 0, y′(0) = 0, y(1) −2y′(1) = 0 13. C y′′ + λy = 0, y(0) −y′(0) = 0, y′(π) = 0 14. C y′′ + λy = 0, y(0) + 2y′(0) = 0, y(π) = 0 15. C y′′ + λy = 0, y′(0) = 0, y(2) −y′(2) = 0 16. C y′′ + λy = 0, y(0) + y′(0) = 0, y(2) + 2y′(2) = 0 17. C y′′ + λy = 0, y(0) + 2y′(0) = 0, y(3) −2y′(3) = 0 18. C y′′ + λy = 0, 3y(0) + y′(0) = 0, 3y(2) −2y′(2) = 0 19. C y′′ + λy = 0, y(0) + 2y′(0) = 0, y(3) −y′(3) = 0 20. C y′′ + λy = 0, 5y(0) + 2y′(0) = 0, 5y(1) −2y′(1) = 0 21. Find the first five eigenvalues of the boundary value problem y′′ + 2y′ + y + λy = 0, y(0) = 0, y′(1) = 0 with errors not greater than 5 × 10−8. State the form of the associated eigenfunctions. In Exercises 22–24 take it as given that {xekx, xe−kx} and {x cos kx, x sinkx} are fundamental sets of solutions of x2y′′ −2xy′ + 2y −k2x2y = 0 and x2y′′ −2xy′ + 2y + k2x2y = 0, respectively. 22. Solve the eigenvalue problem for x2y′′ −2xy′ + 2y + λx2y = 0, y(1) = 0, y(2) = 0. 23. C Find the first five eigenvalues of x2y′′ −2xy′ + 2y + λx2y = 0, y′(1) = 0, y(2) = 0 with errors no greater than 5 × 10−8. State the form of the associated eienfunctions. | Elementary Differential Equations with Boundary Value Problems_Page_708_Chunk3110 |
Section 13.2 Sturm-Liouville Problems 699 24. C Find the first five eigenvalues of x2y′′ −2xy′ + 2y + λx2y = 0, y(1) = 0, y′(2) = 0 with errors no greater than 5 × 10−8. State the form of the associated eienfunctions. 25. Consider the Sturm-Liouville problem y′′ + λy = 0, y(0) = 0, y(L) + δy′(L) = 0. (A) (a) Show that (A) can’t have more than one negative eigenvalue, and find the values of δ for which it has one. (b) Find all values of δ such that λ = 0 is an eigenvalue of (A). (c) Show that λ = k2 with k > 0 is an eigenvalue of (A) if and only if tan kL = −δk. (B) (d) For n = 1, 2, ..., let yn be an eigenfunction associated with λn = k2 n. From Theorem 13.2.4, ym and yn are orthogonal over [0, L] if m ̸= n. Verify this directly. HINT: Integrate by parts twice and use (B). 26. Solve the Sturm-Liouville problem y′′ + λy = 0, y(0) + αy′(0) = 0, y(π) + αy′(π) = 0, where α ̸= 0. 27. L Consider the Sturm-Liouville problem y′′ + λy = 0, y(0) + αy′(0) = 0, y(1) + (α −1)y′(1) = 0, (A) where 0 < α < 1. (a) Show that λ = 0 is an eigenvalue of (A), and find an associated eigenfunction. (b) Show that (A) has a negative eigenvalue, and find the form of an associated eigenfunction. (c) Give a graphical argument to show that (A) has infinitely many positive eigenvalues λ1 < λ2 < · · · < λn < · · ·, and state the form of the associated eigenfunctions. Exercises 28–30 deal with the Sturm–Liouville problem y′′ + λy = 0, αy(0) + βy′(0), ρy(L) + δy′(L) = 0, (SL) where α2 + β2 > 0 and ρ2 + δ2 > 0. 28. Show that λ = 0 is an eigenvalue of (SL) if and only if α(ρL + δ) −βρ = 0. 29. L The point of this exercise is that (SL) can’t have more than two negative eigenvalues. (a) Show that λ is a negative eigenvalue of (SL) if and only if λ = −k2, where k is a positive solution of (αρ −βδk2) sinh kL + k(αδ −βρ) cosh kL. | Elementary Differential Equations with Boundary Value Problems_Page_709_Chunk3111 |
700 Chapter 13 Boundary Value Problems for Second Order Ordinary Differential Equations (b) Suppose αδ −βρ = 0. Show that (SL) has a negative eigenvalue if and only if αρ and βδ are both nonzero. Find the negative eigenvalue and an associated eigenfunction. HINT: Show that in this case ρ = pα and s = qβ, where q ̸= 0. (c) Suppose βρ −αδ ̸= 0. We know from Section 11.1 that (SL) has no negative eigenvalues if αρ = 0 and βδ = 0. Assume that either αρ ̸= 0 or βδ ̸= 0. Then we can rewrite (A) as tanh kL = k(βρ −αδ) αρ −βδk2 . By graphing both sides of this equation on the same axes (there are several possibilitiesfor the right side), show that it has at most two positive solutions, so (SL) has at most two negative eigenvalues. 30. L The point of this exercise is that (SL) has infinitely many positive eigenvalues λ1 < λ2 < · · · < λn < · · ·, and that limn→∞λn = ∞. (a) Show that λ is a positive eigenvalue of (SL) if and only if λ = k2, where k is a positive solution of (αρ + βδk2) sin kL + k(αδ −βρ) cos kL = 0. (A) (b) Suppose αδ −βρ = 0. Show that the positive eigenvalues of (SL) are λn = (nπ/L)2, n = 1, 2, 3, .... HINT: Recall the hint in Exercise 29(b). Now suppose αδ −βρ ̸= 0. From Section 11.1, if αρ = 0 and βδ = 0, then (SL) has the eigenvalues λn = [(2n −1)π/2L]2, n = 1, 2, 3, . . . (why?), so let’s suppose in addition that at least one of the products αρ and βδ is nonzero. Then we can rewrite (A) as tan kL = k(βρ −αδ) αρ −βδk2 . (B) By graphing both sides of this equation on the same axes (there are several possibilities for the right side), convince yourself of the following: (c) If βδ = 0, there’s a positive integer N such that (B) has one solution kn in each of the intervals ((2n −1)π/L, (2n + 1)π/L)) , n = N, N + 1, N + 2, . . ., (C) and either lim n→∞ kn −(2n −1)π 2L = 0 or lim n→∞ kn −(2n + 1)π 2L = 0. (d) If βδ ̸= 0, there’s a positive integer N such that (B) has one solution kn in each of the intervals (C) and lim n→∞ kn −nπ N = 0. 31. The following Sturm–Liouville problems are genera1izations of Problems 1–4 of Section 11.1. Problem 1: (p(x)y′)′ + λr(x)y = 0, y(a) = 0, y(b) = 0 Problem 2: (p(x)y′)′ + λr(x)y = 0, y′(a) = 0, y′(b) = 0 Problem 3: (p(x)y′)′ + λr(x)y = 0, y(a) = 0, y′(b) = 0 Problem 4: (p(x)y′)′ + λr(x)y = 0, y′(a) = 0, y(b) = 0 Prove: Problems 1–4 have no negative eigenvalues. Moreover, λ = 0 is an eigenvalue of Problem 2 with associated eigenfunction y0 = 1, but λ = 0 isn’t an eigenvalue of Problems 1, 3, and 4. HINT: See the proof of Theorem 11.1.1. | Elementary Differential Equations with Boundary Value Problems_Page_710_Chunk3112 |
Section 13.2 Sturm-Liouville Problems 701 32. Show that the eigenvalues of the Sturm–Liouville problem (p(x)y′)′ + λr(x)y = 0, αy(a) + βy′(a) = 0, ρy(b) + δy′(b) are all positive if αβ ≤0, ρδ ≥0, and (αβ)2 + (ρδ)2 > 0. | Elementary Differential Equations with Boundary Value Problems_Page_711_Chunk3113 |
702 Chapter 13 Boundary Value Problems for Second Order Ordinary Differential Equations A BRIEF TABLE OF INTEGRALS Z uα du = uα+1 α + 1 + c, α ̸= −1 Z du u = ln |u| + c Z cos u du = sin u + c Z sinu du = −cos u + c Z tanu du = −ln | cos u| + c Z cot u du = ln| sin u| + c Z sec2 u du = tan u + c Z csc2 u du = −cot u + c Z sec u du = ln| sec u + tan u| + c Z cos2 u du = u 2 + 1 4 sin 2u + c Z sin2 u du = u 2 −1 4 sin 2u + c Z du 1 + u2 du = tan−1 u + c Z du √ 1 −u2 du = sin−1 u + c Z 1 u2 −1 du = 1 2 ln u −1 u + 1 + c Z cosh u du = sinh u + c Z sinhu du = cosh u + c Z u dv = uv − Z v du Z u cos u du = u sinu + cos u + c | Elementary Differential Equations with Boundary Value Problems_Page_712_Chunk3114 |
Section 13.2 Sturm-Liouville Problems 703 Z u sinu du = −u cos u + sin u + c Z ueu du = ueu −eu + c Z eλu cos ωu du = eλu(λ cos ωu + ω sinωu) λ2 + ω2 + c Z eλu sin ωu du = eλu(λ sin ωu −ω cos ωu) λ2 + ω2 + c Z ln|u| du = u ln|u| −u + c Z u ln|u| du = u2 ln |u| 2 −u2 4 + c Z cos ω1u cos ω2u du = sin(ω1 + ω2)u 2(ω1 + ω2) + sin(ω1 −ω2)u 2(ω1 −ω2) + c (ω1 ̸= ±ω2) Z sinω1u sinω2u du = −sin(ω1 + ω2)u 2(ω1 + ω2) + sin(ω1 −ω2)u 2(ω1 −ω2) + c (ω1 ̸= ±ω2) Z sinω1u cos ω2u du = −cos(ω1 + ω2)u 2(ω1 + ω2) −cos(ω1 −ω2)u 2(ω1 −ω2) + c (ω1 ̸= ±ω2) | Elementary Differential Equations with Boundary Value Problems_Page_713_Chunk3115 |
Answers to Selected Exercises Section 1.2 Answers, pp. 14–15 1.2.1 (p. 14) (a) 3 (b) 2 (c) 1 (d) 2 1.2.3 (p. 14) (a) y = −x2 2 + c (b) y = x cos x −sin x + c (c) y = x2 2 lnx −x2 4 + c (d) y = −x cos x + 2 sin x + c1 + c2x (e) y = (2x −4)ex + c1 + c2x (f) y = x3 3 −sinx + ex + c1 + c2x (g) y = sin x + c1 + c2x + c3x2 (h) y = −x5 60 + ex + c1 + c2x + c3x2 (i) y = 7 64e4x + c1 + c2x + c3x2 1.2.4 (p. 14) (a) y = −(x −1)ex (b) y = 1 −1 2 cos x2 (c) y = 3 −ln( √ 2 cos x) (d) y = −47 15 −37 5 (x −2) + x5 30 (e) y = 1 4xe2x −1 4e2x + 29 4 (f) y = x sinx + 2 cos x −3x −1 (g) y = (x2 −6x + 12)ex + x2 2 −8x −11 (h) y = x3 3 + cos 2x 6 + 7 4x2 −6x + 7 8 (i) y = x4 12 + x3 6 + 1 2(x −2)2 −26 3 (x −2) −5 3 1.2.7 (p. 15) (a) 576 ft (b) 10 s 1.2.8 (p. 15) (b) y = 0 1.2.10 (p. 15) (a) (−2c−2, ∞) (−∞, ∞) 705 | Elementary Differential Equations with Boundary Value Problems_Page_714_Chunk3116 |
Answers to Selected Exercises 705 Section 2.1 Answers, pp. 41–44 2.1.1 (p. 41) y = e−ax 2.1.2 (p. 41) y = ce−x3 2.1.3 (p. 41) y = ce−(ln x)2/2 2.1.4 (p. 41) y = c x3 2.1.5 (p. 41) y = ce1/x 2.1.6 (p. 41) y = e−(x−1) x 2.1.7 (p. 41) y = e x lnx 2.1.8 (p. 41) y = π x sin x 2.1.9 (p. 41) y = 2(1 + x2) 2.1.10 (p. 41) y = 3x−k 2.1.11 (p. 41) y = c(cos kx)1/k 2.1.12 (p. 41) y = 1 3 + ce−3x 2.1.13 (p. 41) y = 2 x + c xex 2.1.14 (p. 41) y = e−x2 x2 2 + c 2.1.15(p. 41) y = −e−x + c 1 + x2 2.1.16(p. 42) y = 7 ln |x| x + 3 2x + c x 2.1.17 (p. 42) y = (x −1)−4(ln |x −1| −cos x + c) 2.1.18 (p. 42) y = e−x2 x3 4 + c x 2.1.19 (p. 42) y = 2 ln |x| x2 + 1 2 + c x2 2.1.20 (p. 42) y = (x+c) cos x 2.1.21 (p. 42) y = c −cos x (1 + x)2 2.1.22 (p. 42) y = −1 2 (x −2)3 (x −1) + c(x −2)5 (x −1) 2.1.23 (p. 42) y = (x + c)e−sin2 x 2.1.24 (p. 42) y = ex x2 −ex x3 + c x2 . y = e3x −e−7x 10 2.1.26 (p. 42) 2x + 1 (1 + x2)2 2.1.27 (p. 42) y = 1 x2 ln 1 + x2 2 2.1.29(p. 42) y = 2 ln |x| x + x 2 −1 2x 2.1.28 (p. 42) y = 1 2(sin x + csc x) 2.1.29 (p. 42) y = 2 ln |x| x + x 2 −1 2x 2.1.30 (p. 42) y = (x −1)−3 [ln(1 −x) −cos x] 2.1.31 (p. 42) y = 2x2 + 1 x2 (0, ∞) 2.1.32(p. 42) y = x2(1−ln x) 2.1.33(p. 42) y = 1 2 + 5 2e−x2 2.1.34 (p. 42) y = ln |x −1| + tan x + 1 (x −1)3 2.1.35 (p. 42) y = ln|x| + x2 + 1 (x + 2)4 2.1.36 (p. 42) y = (x2 −1) 1 2 ln|x2 −1| −4 2.1.37 (p. 42) y = −(x2 −5) | Elementary Differential Equations with Boundary Value Problems_Page_715_Chunk3117 |
706 Answers to Selected Exercises 2.1.43 (p. 43) G = r λ + G0 −r λ e−λt limt→∞G(t) = r λ 2.1.45 (p. 43) (a) y = y0e−a(x−x0) + e−ax Z x x0 eatf(t) dt 2.1.48 (p. 44) (a) y = tan−1 1 3 + ce3x (b) y = ± ln 1 x + c x2 1/2 (c) y = exp x2 + c x2 (d) y = −1 + x c + 3 ln|x| Section 2.2 Answers, pp. 52–55 2.2.1 (p. 52) y = 2 ± p 2(x3 + x2 + x + c) 2.2.2 (p. 52) ln(| siny|) = cos x + c; y ≡kπ, k = integer 2.2.3 (p. 52) y = c x −c y ≡−1 2.2.4 (p. 52) (ln y)2 2 = −x3 3 + c 2.2.5 (p. 52) y3 + 3 sin y + ln|y| + ln(1 + x2) + tan−1 x = c; y ≡0 2.2.6 (p. 52) y = ± 1 + x 1 + cx 2!1/2 ; y ≡±1 2.2.7 (p. 52) y = tan x3 3 + c 2.2.8 (p. 52) y = c √ 1 + x2 2.2.9(p. 52) y = 2 −ce(x−1)2/2 1 −ce(x−1)2/2 ; y ≡1 2.2.10 (p. 52) y = 1 + | Elementary Differential Equations with Boundary Value Problems_Page_716_Chunk3118 |
Answers to Selected Exercises 707 2.2.24 (p. 53) y = x + c 1 −cx 2.2.25 (p. 53) y = −x cos c + √ 1 −x2 sin c; y ≡1; y ≡−1 2.2.26 (p. 53) y = −x + 3π/2 2.2.28 (p. 53) P = P0 αP0 + (1 −αP0)e−at ; limt→∞P (t) = 1/α 2.2.29 (p. 53) I = SI0 I0 + (S −I0)e−rSt 2.2.30 (p. 53) If q = rS then I = I0 1 + rI0t and limt→∞I(t) = 0. If q ̸= Rs, then I = αI0 I0 + (α −I0)e−rαt . If q < rs, then limt→∞I(t) = α = S −q r if q > rS, then limt→∞I(t) = 0 2.2.34 (p. 55) f = ap, where a=constant 2.2.35 (p. 55) y = e−x | Elementary Differential Equations with Boundary Value Problems_Page_717_Chunk3119 |
708 Answers to Selected Exercises 2.3.12 (p. 61) (a), (b) all (x0, y0) such that x0 + y0 > 0 2.3.13 (p. 61) (a), (b) all (x0, y0) with x0 ̸= 1, y0 ̸= (2k + 1)π 2 (k = integer) 2.3.16 (p. 61) y = 3 5x + 1 5/3 , −∞< x < ∞, is a solution. Also, y = ( 0, −∞< x ≤−5 3 | Elementary Differential Equations with Boundary Value Problems_Page_718_Chunk3120 |
Answers to Selected Exercises 709 is a solution of (A) on (−∞, ∞). If α ≥0, then y = 1 + (x2 −α2)3/2, −∞< x < −α, 1, −α ≤x ≤ √ 5, 1 −(x2 −5)3/2, √ 5 < x < ∞, and y = 1 −(x2 −α2)3/2, −∞< x < −α, 1, −α ≤x ≤ √ 5, 1 −(x2 −5)3/2, √ 5 < x < ∞, are also solutions of (A) on (−∞, ∞). Section 2.4 Answers, pp. 68–72 2.4.1 (p. 68) y = 1 1 −cex 2.4.2 (p. 68) y = x2/7(c −ln|x|)1/7 2.4.3 (p. 68) y = e2/x(c −1/x)2 2.4.4 (p. 68) y = ± √2x + c 1 + x2 2.4.5 (p. 68) y = ±(1 −x2 + ce−x2)−1/2 2.4.6 (p. 68) y = x 3(1 −x) + ce−x 1/3 2.4.7(p. 69) y = 2 √ 2 √1 −4x 2.4.8 (p. 69) y = 1 −3 2e−(x2−1)/4 −2 2.4.9 (p. 69) y = 1 x(11 −3x)1/3 2.4.10 (p. 69) y = (2ex −1)2 2.4.11 (p. 69) y = (2e12x −1 −12x)1/3 2.4.12 (p. 69) y = 5x 2(1 + 4x5) 1/2 2.4.13 (p. 69) y = (4ex/2 −x −2)2 2.4.14 (p. 69) P = P0eat 1 + aP0 R t 0 α(τ)eaτ dτ ; limt→∞P (t) = ∞ if L = 0, 0 if L = ∞, 1/aL if 0 < L < ∞. 2.4.15 (p. 69) y = x(ln |x| + c) 2.4.16 (p. 69) y = cx2 1 −cx y = −x 2.4.17 (p. 69) y = ±x(4 ln |x| + c)1/4 2.4.18 (p. 69) y = x sin−1(ln|x| + c) 2.4.19 (p. 69) y = x tan(ln |x| + c) 2.4.20 (p. 69) y = ±x √ cx2 −1 2.4.21 (p. 70) y = ±x ln(ln |x| + c) 2.4.22 (p. 70) y = − 2x 2 ln |x| + 1 2.4.23 (p. 70) y = x(3 lnx + 27)1/3 2.4.24 (p. 70) y = 1 x 9 −x4 2 1/2 2.4.25 (p. 70) y = −x | Elementary Differential Equations with Boundary Value Problems_Page_719_Chunk3121 |
710 Answers to Selected Exercises 2.4.26 (p. 70) y = −x(4x −3) (2x −3) 2.4.27(p. 70) y = x √ 4x6 −1 2.4.28 (p. 70) tan−1 y x −1 2 ln(x2 + y2) = c 2.4.29 (p. 70) (x + y) ln |x| + y(1 −ln |y|) + cx = 0 2.4.30 (p. 70) (y + x)3 = 3x3(ln|x| + c) 2.4.31 (p. 70) (y + x) = c(y −x)3; y = x; y = −x 2.4.32 (p. 70) y2(y −3x) = c; y ≡0; y = 3x 2.4.33 (p. 70) (x−y)3(x+y) = cy2x4; y = 0; y = x; y = −x 2.4.34 (p. 70) y x + y3 x3 = ln |x|+c 2.4.40 (p. 71) Choose X0 and Y0 so that aX0 + bY0 = α cX0 + dY0 = β. 2.4.41 (p. 72) (y + 2x + 1)4(2y −6x −3) = c; y = 3x + 3/2; y = −2x −1 2.4.42 (p. 72) (y + x −1)(y −x −5)3 = c; y = x + 5; y = −x + 1 2.4.43 (p. 72) ln|y −x −6| −2(x + 2) y −x −6 = c; y = x + 6 2.4.44 (p. 72) (y1 = x1/3) y = x1/3(ln |x| + c)1/3 2.4.45 (p. 72) y1 = x3; y = ±x3√ cx6 −1 2.4.46 (p. 72) y1 = x2; y = x2(1 + cx4) 1 −cx4 y = −x2 2.4.47 (p. 72) y1 = ex; y = −ex(1 −2cex) 1 −cex ; y = −2ex 2.4.48 (p. 72) y1 = tan x; y = tan x tan(ln | tanx| + c) 2.4.49 (p. 72) y1 = ln x; y = 2 lnx | Elementary Differential Equations with Boundary Value Problems_Page_720_Chunk3122 |
Answers to Selected Exercises 711 2.4.57 (p. 72) y = ex − 1 1 + ce−x 2.4.58 (p. 72) y = 1 − 1 x(1 −cx) 2.4.59 (p. 72) y = x − 2x x2 + c | Elementary Differential Equations with Boundary Value Problems_Page_721_Chunk3123 |
712 Answers to Selected Exercises Section 2.5 Answers, pp. 79–82 2.5.1 (p. 79) 2x3y2 = c 2.5.2 (p. 79) 3y sinx + 2x2ex + 3y = c 2.5.3 (p. 79) Not exact 2.5.4 (p. 79) x2 −2xy2 + 4y3 = c 2.5.5 (p. 79) x + y = c 2.5.6 (p. 79) Not exact 2.5.7 (p. 79) 2y2 cos x + 3xy3 −x2 = c 2.5.8 (p. 79) Not exact 2.5.9 (p. 79) x3+x2y+4xy2+9y2 = c 2.5.10 (p. 79) Not exact 2.5.11 (p. 79) ln |xy|+x2+y2 = c 2.5.12 (p. 79) Not exact 2.5.13 (p. 79) x2 + y2 = c 2.5.14 (p. 79) x2y2ex + 2y + 3x2 = c 2.5.15 (p. 79) x3ex2+y −4y3 + 2x2 = c 2.5.16 (p. 79) x4exy + 3xy = c 2.5.17 (p. 79) x3 cos xy + 4y2 + 2x2 = c 2.5.18 (p. 79) y = x + √ 2x2 + 3x −1 x2 2.5.19 (p. 79) y = sin x − r 1 −tan x 2 2.5.20 (p. 79) y = ex −1 ex + 1 1/3 2.5.21 (p. 79) y = 1 + 2 tanx 2.5.22 (p. 79) y = x2 −x + 6 (x + 2)(x −3) 2.5.23 (p. 80) 7x2 2 + 4xy + 3y2 2 = c 2.5.24 (p. 80) (x4y2 + 1)ex + y2 = c 2.5.29 (p. 81) (a) M(x, y) = 2xy + f(x) (b) M(x, y) = 2(sin x + x cos x)(y sin y + cos y) + f(x) (c) M(x, y) = yex −ey cos x + f(x) 2.5.30 (p. 81) (a) N(x, y) = x4y 2 + x2 + 6xy + g(y) (b) N(x, y) = x y + 2y sin x + g(y) (c) N(x, y) = x(sin y + y cos y) + g(y) 2.5.33 (p. 81) B = C 2.5.34 (p. 81) B = 2D, E = 2C 2.5.37 (p. 82) (a) 2x2 + x4y4 + y2 = c (b) x3 + 3xy2 = c (c) x3 + y2 + 2xy = c 2.5.38 (p. 82) y = −1 −1 x2 2.5.39 (p. 82) y = x3 −3(x2 + 1) + √ 9x4 + 34x2 + 21 2 ! | Elementary Differential Equations with Boundary Value Problems_Page_722_Chunk3124 |
Answers to Selected Exercises 713 2.5.40 (p. 82) y = −e−x2 2x + √ 9 −5x2 3 ! . 2.5.44 (p. 82) (a) G(x, y) = 2xy + c (b) G(x, y) = ex sin y + c (c) G(x, y) = 3x2y −y3 + c (d) G(x, y) = −sin x sinhy + c (e) G(x, y) = cos x sinh y + c Section 2.6 Answers, pp. 91–93 2.6.3 (p. 91) µ(x) = 1/x2; y = cx and µ(y) = 1/y2; x = cy 2.6.4 (p. 91) µ(x) = x−3/2; x3/2y = c 2.6.5 (p. 91) µ(y) = 1/y3; y3e2x = c 2.6.6 (p. 91) µ(x) = e5x/2; e5x/2(xy + 1) = c 2.6.7 (p. 92) µ(x) = ex; ex(xy + y + x) = c 2.6.8 (p. 92) µ(x) = x; x2y2(9x+4y) = c 2.6.9 (p. 92) µ(y) = y2; y3(3x2y +2x+1) = c 2.6.10 (p. 92) µ(y) = yey; ey(xy3 + 1) = c 2.6.11 (p. 92) µ(y) = y2; y3(3x4 + 8x3y + y) = c 2.6.12 (p. 92) µ(x) = xex; x2y(x + 1)ex = c 2.6.13 (p. 92) µ(x) = (x3 −1)−4/3; xy(x3 −1)−1/3 = c and x ≡1 2.6.14 (p. 92) µ(y) = ey; ey(sin x cos y+y−1) = c 2.6.15 (p. 92) µ(y) = e−y2; xye−y2(x+y) = c 2.6.16 (p. 92) xy sin y = c and y = kπ (k = integer) 2.6.17 (p. 92) µ(x, y) = x4y3; x5y4 ln x = c 2.6.18 (p. 92) µ(x, y) = 1/xy; |x|α|y|βeγxeδy = c and x ≡0, y ≡0 2.6.19 (p. 92) µ(x, y) = x−2y−3; 3x2y2 + y = 1 + cxy2 and x ≡0, y ≡0 2.6.20 (p. 92) µ(x, y) = x−2y−1; −2 x + y3 + 3 ln |y| = c and x ≡0, y ≡0 2.6.21 (p. 92) µ(x, y) = eaxeby; eaxeby cos xy = c | Elementary Differential Equations with Boundary Value Problems_Page_723_Chunk3125 |
714 Answers to Selected Exercises 2.6.22 (p. 92) µ(x, y) = x−4y−3 (and others) xy = c 2.6.23 (p. 92) µ(x, y) = xey; x2yey sin x = c 2.6.24 (p. 92) µ(x) = 1/x2; x3y3 3 −y x = c 2.6.25 (p. 92) µ(x) = x + 1; y(x + 1)2(x + y) = c 2.6.26 (p. 92) µ(x, y) = x2y2; x3y3(3x + 2y2) = c 2.6.27 (p. 92) µ(x, y) = x−2y−2; 3x2y = cxy + 2 and x ≡0, y ≡0 Section 3.1 Answers, pp. 106–108 3.1.1 (p. 106) y1 = 1.450000000, y2 = 2.085625000, y3 = 3.079099746 3.1.2 (p. 106) y1 = 1.200000000, y2 = 1.440415946, y3 = 1.729880994 3.1.3 (p. 106) y1 = 1.900000000, y2 = 1.781375000, y3 = 1.646612970 3.1.4 (p. 106) y1 = 2.962500000, y2 = 2.922635828, y3 = 2.880205639 3.1.5 (p. 106) y1 = 2.513274123, y2 = 1.814517822, y3 = 1.216364496 3.1.6 (p. 106) x h = 0.1 h = 0.05 h = 0.025 Exact 1.0 48.298147362 51.492825643 53.076673685 54.647937102 3.1.7 (p. 106) x h = 0.1 h = 0.05 h = 0.025 Exact 2.0 1.390242009 1.370996758 1.361921132 1.353193719 3.1.8 (p. 107) x h = 0.05 h = 0.025 h = 0.0125 Exact 1.50 7.886170437 8.852463793 9.548039907 10.500000000 3.1.9 (p. 107) x h = 0.1 h = 0.05 h = 0.025 h = 0.1 h = 0.05 h = 0.025 3.0 1.469458241 1.462514486 1.459217010 0.3210 0.1537 0.0753 Approximate Solutions Residuals 3.1.10 (p. 107) x h = 0.1 h = 0.05 h = 0.025 h = 0.1 h = 0.05 h = 0.025 2.0 0.473456737 0.483227470 0.487986391 -0.3129 -0.1563 -0.0781 Approximate Solutions Residuals 3.1.11 (p. 107) x h = 0.1 h = 0.05 h = 0.025 “Exact” 1.0 0.691066797 0.676269516 0.668327471 0.659957689 | Elementary Differential Equations with Boundary Value Problems_Page_724_Chunk3126 |
Answers to Selected Exercises 715 3.1.12 (p. 108) x h = 0.1 h = 0.05 h = 0.025 “Exact” 2.0 -0.772381768 -0.761510960 -0.756179726 -0.750912371 3.1.13 (p. 108) Euler’s method x h = 0.1 h = 0.05 h = 0.025 Exact 1.0 0.538871178 0.593002325 0.620131525 0.647231889 Euler semilinear method x h = 0.1 h = 0.05 h = 0.025 Exact 1.0 0.647231889 0.647231889 0.647231889 0.647231889 Applying variation of parameters to the given initial value problem yields y = ue−3x, where (A) u′ = 7, u(0) = 6. Since u′′ = 0, Euler’s method yields the exact solution of (A). Therefore the Euler semilinear method produces the exact solution of the given problem . 3.1.14 (p. 108) Euler’s method x h = 0.1 h = 0.05 h = 0.025 “Exact” 3.0 12.804226135 13.912944662 14.559623055 15.282004826 Euler semilinear method x h = 0.1 h = 0.05 h = 0.025 “Exact” 3.0 15.354122287 15.317257705 15.299429421 15.282004826 3.1.15 (p. 108) Euler’s method x h = 0.2 h = 0.1 h = 0.05 “Exact” 2.0 0.867565004 0.885719263 0.895024772 0.904276722 Euler semilinear method x h = 0.2 h = 0.1 h = 0.05 “Exact” 2.0 0.569670789 0.720861858 0.808438261 0.904276722 3.1.16 (p. 108) Euler’s method x h = 0.2 h = 0.1 h = 0.05 “Exact” 3.0 0.922094379 0.945604800 0.956752868 0.967523153 Euler semilinear method x h = 0.2 h = 0.1 h = 0.05 “Exact” 3.0 0.993954754 0.980751307 0.974140320 0.967523153 | Elementary Differential Equations with Boundary Value Problems_Page_725_Chunk3127 |
716 Answers to Selected Exercises 3.1.17 (p. 108) Euler’s method x h = 0.0500 h = 0.0250 h = 0.0125 “Exact” 1.50 0.319892131 0.330797109 0.337020123 0.343780513 Euler semilinear method x h = 0.0500 h = 0.0250 h = 0.0125 “Exact” 1.50 0.305596953 0.323340268 0.333204519 0.343780513 3.1.18 (p. 108) Euler’s method x h = 0.2 h = 0.1 h = 0.05 “Exact” 2.0 0.754572560 0.743869878 0.738303914 0.732638628 Euler semilinear method x h = 0.2 h = 0.1 h = 0.05 “Exact” 2.0 0.722610454 0.727742966 0.730220211 0.732638628 3.1.19 (p. 108) Euler’s method x h = 0.0500 h = 0.0250 h = 0.0125 “Exact” 1.50 2.175959970 2.210259554 2.227207500 2.244023982 Euler semilinear method x h = 0.0500 h = 0.0250 h = 0.0125 “Exact” 1.50 2.117953342 2.179844585 2.211647904 2.244023982 3.1.20 (p. 108) Euler’s method x h = 0.1 h = 0.05 h = 0.025 “Exact” 1.0 0.032105117 0.043997045 0.050159310 0.056415515 Euler semilinear method x h = 0.1 h = 0.05 h = 0.025 “Exact” 1.0 0.056020154 0.056243980 0.056336491 0.056415515 3.1.21 (p. 108) Euler’s method x h = 0.1 h = 0.05 h = 0.025 “Exact” 1.0 28.987816656 38.426957516 45.367269688 54.729594761 Euler semilinear method x h = 0.1 h = 0.05 h = 0.025 “Exact” 1.0 54.709134946 54.724150485 54.728228015 54.729594761 3.1.22 (p. 108) Euler’s method x h = 0.1 h = 0.05 h = 0.025 “Exact” 3.0 1.361427907 1.361320824 1.361332589 1.361383810 Euler semilinear method x h = 0.1 h = 0.05 h = 0.025 “Exact” 3.0 1.291345518 1.326535737 1.344004102 1.361383810 Section 3.2 Answers, pp. 116–108 | Elementary Differential Equations with Boundary Value Problems_Page_726_Chunk3128 |
Answers to Selected Exercises 717 3.2.1 (p. 116) y1 = 1.542812500, y2 = 2.421622101, y3 = 4.208020541 3.2.2 (p. 116) y1 = 1.220207973, y2 = 1.489578775 y3 = 1.819337186 3.2.3 (p. 116) y1 = 1.890687500, y2 = 1.763784003, y3 = 1.622698378 3.2.4 (p. 116) y1 = 2.961317914 y2 = 2.920132727 y3 = 2.876213748. 3.2.5 (p. 116) y1 = 2.478055238, y2 = 1.844042564, y3 = 1.313882333 3.2.6 (p. 116) x h = 0.1 h = 0.05 h = 0.025 Exact 1.0 56.134480009 55.003390448 54.734674836 54.647937102 3.2.7 (p. 116) x h = 0.1 h = 0.05 h = 0.025 Exact 2.0 1.353501839 1.353288493 1.353219485 1.353193719 3.2.8 (p. 117) x h = 0.05 h = 0.025 h = 0.0125 Exact 1.50 10.141969585 10.396770409 10.472502111 10.500000000 3.2.9 (p. 117) x h = 0.1 h = 0.05 h = 0.025 h = 0.1 h = 0.05 h = 0.025 3.0 1.455674816 1.455935127 1.456001289 -0.00818 -0.00207 -0.000518 Approximate Solutions Residuals 3.2.10 (p. 117) x h = 0.1 h = 0.05 h = 0.025 h = 0.1 h = 0.05 h = 0.025 2.0 0.492862999 0.492709931 0.492674855 0.00335 0.000777 0.000187 Approximate Solutions Residuals 3.2.11 (p. 117) x h = 0.1 h = 0.05 h = 0.025 “Exact" 1.0 0.660268159 0.660028505 0.659974464 0.659957689 3.2.12 (p. 118) x h = 0.1 h = 0.05 h = 0.025 “Exact" 2.0 -0.749751364 -0.750637632 -0.750845571 -0.750912371 3.2.13 (p. 118) Applying variation of parameters to the given initial value problem y = ue−3x, where (A) u′ = 1 −2x, u(0) = 2. Since u′′′ = 0, the improved Euler method yields the exact solution of (A). Therefore the improved Euler semilinear method produces the exact solution of the given problem. Improved Euler method x h = 0.1 h = 0.05 h = 0.025 Exact 1.0 0.105660401 0.100924399 0.099893685 0.099574137 Improved Euler semilinear method x h = 0.1 h = 0.05 h = 0.025 Exact 1.0 0.099574137 0.099574137 0.099574137 0.099574137 3.2.14 (p. 118) Improved Euler method x h = 0.1 h = 0.05 h = 0.025 “Exact" 3.0 15.107600968 15.234856000 15.269755072 15.282004826 Improved Euler semilinear method x h = 0.1 h = 0.05 h = 0.025 “Exact" 3.0 15.285231726 15.282812424 15.282206780 15.282004826 | Elementary Differential Equations with Boundary Value Problems_Page_727_Chunk3129 |
718 Answers to Selected Exercises 3.2.15 (p. 118) Improved Euler method x h = 0.2 h = 0.1 h = 0.05 “Exact" 2.0 0.924335375 0.907866081 0.905058201 0.904276722 Improved Euler semilinear method x h = 0.2 h = 0.1 h = 0.05 “Exact" 2.0 0.969670789 0.920861858 0.908438261 0.904276722 3.2.16 (p. 118) Improved Euler method x h = 0.2 h = 0.1 h = 0.05 “Exact" 3.0 0.967473721 0.967510790 0.967520062 0.967523153 Improved Euler semilinear method x h = 0.2 h = 0.1 h = 0.05 “Exact" 3.0 0.967473721 0.967510790 0.967520062 0.967523153 3.2.17 (p. 118) Improved Euler method x h = 0.0500 h = 0.0250 h = 0.0125 “Exact" 1.50 0.349176060 0.345171664 0.344131282 0.343780513 Improved Euler semilinear method x h = 0.0500 h = 0.0250 h = 0.0125 “Exact" 1.50 0.349350206 0.345216894 0.344142832 0.343780513 3.2.18 (p. 118) Improved Euler method x h = 0.2 h = 0.1 h = 0.05 “Exact" 2.0 0.732679223 0.732721613 0.732667905 0.732638628 Improved Euler semilinear method x h = 0.2 h = 0.1 h = 0.05 “Exact" 2.0 0.732166678 0.732521078 0.732609267 0.732638628 3.2.19 (p. 118) Improved Euler method x h = 0.0500 h = 0.0250 h = 0.0125 “Exact" 1.50 2.247880315 2.244975181 2.244260143 2.244023982 Improved Euler semilinear method x h = 0.0500 h = 0.0250 h = 0.0125 “Exact" 1.50 2.248603585 2.245169707 2.244310465 2.244023982 3.2.20 (p. 118) Improved Euler method x h = 0.1 h = 0.05 h = 0.025 “Exact" 1.0 0.059071894 0.056999028 0.056553023 0.056415515 Improved Euler semilinear method x h = 0.1 h = 0.05 h = 0.025 “Exact" 1.0 0.056295914 0.056385765 0.056408124 0.056415515 | Elementary Differential Equations with Boundary Value Problems_Page_728_Chunk3130 |
Answers to Selected Exercises 719 3.2.21 (p. 118) Improved Euler method x h = 0.1 h = 0.05 h = 0.025 “Exact" 1.0 50.534556346 53.483947013 54.391544440 54.729594761 Improved Euler semilinear method x h = 0.1 h = 0.05 h = 0.025 “Exact" 1.0 54.709041434 54.724083572 54.728191366 54.729594761 3.2.22 (p. 118) Improved Euler method x h = 0.1 h = 0.05 h = 0.025 “Exact" 3.0 1.361395309 1.361379259 1.361382239 1.361383810 Improved Euler semilinear method x h = 0.1 h = 0.05 h = 0.025 “Exact" 3.0 1.375699933 1.364730937 1.362193997 1.361383810 3.2.23 (p. 118) x h = 0.1 h = 0.05 h = 0.025 Exact 2.0 1.349489056 1.352345900 1.352990822 1.353193719 3.2.24 (p. 118) x h = 0.1 h = 0.05 h = 0.025 Exact 2.0 1.350890736 1.352667599 1.353067951 1.353193719 3.2.25 (p. 118) x h = 0.05 h = 0.025 h = 0.0125 Exact 1.50 10.133021311 10.391655098 10.470731411 10.500000000 3.2.26 (p. 118) x h = 0.05 h = 0.025 h = 0.0125 Exact 1.50 10.136329642 10.393419681 10.470731411 10.500000000 3.2.27 (p. 118) x h = 0.1 h = 0.05 h = 0.025 “Exact" 1.0 0.660846835 0.660189749 0.660016904 0.659957689 3.2.28 (p. 119) x h = 0.1 h = 0.05 h = 0.025 “Exact" 1.0 0.660658411 0.660136630 0.660002840 0.659957689 3.2.29 (p. 119) x h = 0.1 h = 0.05 h = 0.025 “Exact" 2.0 -0.750626284 -0.750844513 -0.750895864 -0.751331499 3.2.30 (p. 119) x h = 0.1 h = 0.05 h = 0.025 “Exact" 2.0 -0.750335016 -0.750775571 -0.750879100 -0.751331499 Section 3.3 Answers, pp. 124–127 3.3.1 (p. 124) y1 = 1.550598190, y2 = 2.469649729 3.3.2 (p. 124) y1 = 1.221551366, y2 = 1.492920208 3.3.3 (p. 124) y1 = 1.890339767, y2 = 1.763094323 3.3.4 (p. 124) y1 = 2.961316248 y2 = 2.920128958. 3.3.5 (p. 124) y1 = 2.475605264, y2 = 1.825992433 3.3.6 (p. 124) x h = 0.1 h = 0.05 h = 0.025 Exact 1.0 54.654509699 54.648344019 54.647962328 54.647937102 | Elementary Differential Equations with Boundary Value Problems_Page_729_Chunk3131 |
720 Answers to Selected Exercises 3.3.7 (p. 124) x h = 0.1 h = 0.05 h = 0.025 Exact 2.0 1.353191745 1.353193606 1.353193712 1.353193719 3.3.8 (p. 125) x h = 0.05 h = 0.025 h = 0.0125 Exact 1.50 10.498658198 10.499906266 10.499993820 10.500000000 3.3.9 (p. 125) x h = 0.1 h = 0.05 h = 0.025 h = 0.1 h = 0.05 h = 0.025 3.0 1.456023907 1.456023403 1.456023379 0.0000124 0.000000611 0.0000000333 Approximate Solutions Residuals 3.3.10 (p. 125) x h = 0.1 h = 0.05 h = 0.025 h = 0.1 h = 0.05 h = 0.025 2.0 0.492663789 0.492663738 0.492663736 0.000000902 0.0000000508 0.00000000302 Approximate Solutions Residuals 3.3.11 (p. 125) x h = 0.1 h = 0.05 h = 0.025 “Exact" 1.0 0.659957046 0.659957646 0.659957686 0.659957689 3.3.12 (p. 126) x h = 0.1 h = 0.05 h = 0.025 “Exact" 2.0 -0.750911103 -0.750912294 -0.750912367 -0.750912371 3.3.13 (p. 126) Applying variation of parameters to the given initial value problem yields y = ue−3x, where (A) u′ = 1 −4x + 3x2 −4x3, u(0) = −3. Since u(5) = 0, the Runge-Kutta method yields the exact solution of (A). Therefore the Euler semilinear method produces the exact solution of the given problem. Runge-Kutta method x h = 0.1 h = 0.05 h = 0.025 Exact 0.0 -3.000000000 -3.000000000 -3.000000000 -3.000000000 0.1 -2.162598011 -2.162526572 -2.162522707 -2.162522468 0.2 -1.577172164 -1.577070939 -1.577065457 -1.577065117 0.3 -1.163350794 -1.163242678 -1.163236817 -1.163236453 0.4 -0.868030294 -0.867927182 -0.867921588 -0.867921241 0.5 -0.655542739 -0.655450183 -0.655445157 -0.655444845 0.6 -0.501535352 -0.501455325 -0.501450977 -0.501450707 0.7 -0.389127673 -0.389060213 -0.389056546 -0.389056318 0.8 -0.306468018 -0.306412184 -0.306409148 -0.306408959 0.9 -0.245153433 -0.245107859 -0.245105379 -0.245105226 1.0 -0.199187198 -0.199150401 -0.199148398 -0.199148273 Runge-Kutta semilinear method x h = 0.1 h = 0.05 h = 0.025 Exact 0.0 -3.000000000 -3.000000000 -3.000000000 -3.000000000 0.1 -2.162522468 -2.162522468 -2.162522468 -2.162522468 0.2 -1.577065117 -1.577065117 -1.577065117 -1.577065117 0.3 -1.163236453 -1.163236453 -1.163236453 -1.163236453 0.4 -0.867921241 -0.867921241 -0.867921241 -0.867921241 0.5 -0.655444845 -0.655444845 -0.655444845 -0.655444845 0.6 -0.501450707 -0.501450707 -0.501450707 -0.501450707 0.7 -0.389056318 -0.389056318 -0.389056318 -0.389056318 0.8 -0.306408959 -0.306408959 -0.306408959 -0.306408959 0.9 -0.245105226 -0.245105226 -0.245105226 -0.245105226 1.0 -0.199148273 -0.199148273 -0.199148273 -0.199148273 | Elementary Differential Equations with Boundary Value Problems_Page_730_Chunk3132 |
Answers to Selected Exercises 721 3.3.14 (p. 126) Runge-Kutta method x h = 0.1 h = 0.05 h = 0.025 “Exact" 3.0 15.281660036 15.281981407 15.282003300 15.282004826 Runge-Kutta semilinear method x h = 0.1 h = 0.05 h = 0.025 “Exact" 3.0 15.282005990 15.282004899 15.282004831 15.282004826 3.3.15 (p. 126) Runge-Kutta method x h = 0.2 h = 0.1 h = 0.05 “Exact" 2.0 0.904678156 0.904295772 0.904277759 0.904276722 Runge-Kutta semilinear method x h = 0.2 h = 0.1 h = 0.05 “Exact" 2.0 0.904592215 0.904297062 0.904278004 0.904276722 3.3.16 (p. 126) Runge-Kutta method x h = 0.2 h = 0.1 h = 0.05 “Exact" 3.0 0.967523147 0.967523152 0.967523153 0.967523153 Runge-Kutta semilinear method x h = 0.2 h = 0.1 h = 0.05 “Exact" 3.0 0.967523147 0.967523152 0.967523153 0.967523153 3.3.17 (p. 126) Runge-Kutta method x h = 0.0500 h = 0.0250 h = 0.0125 “Exact" 1.50 0.343839158 0.343784814 0.343780796 0.343780513 Runge-Kutta semilinear method x h = 0.0500 h = 0.0250 h = 0.0125 “Exact" 1.00 0.000000000 0.000000000 0.000000000 0.000000000 1.05 0.028121022 0.028121010 0.028121010 0.028121010 1.10 0.055393494 0.055393466 0.055393465 0.055393464 1.15 0.082164048 0.082163994 0.082163990 0.082163990 1.20 0.108862698 0.108862597 0.108862591 0.108862590 1.25 0.136058715 0.136058528 0.136058517 0.136058516 1.30 0.164564862 0.164564496 0.164564473 0.164564471 1.35 0.195651074 0.195650271 0.195650219 0.195650216 1.40 0.231542288 0.231540164 0.231540027 0.231540017 1.45 0.276818775 0.276811011 0.276810491 0.276810456 1.50 0.343839124 0.343784811 0.343780796 0.343780513 3.3.18 (p. 126) Runge-Kutta method x h = 0.2 h = 0.1 h = 0.05 “Exact" 2.0 0.732633229 0.732638318 0.732638609 0.732638628 Runge-Kutta semilinear method x h = 0.2 h = 0.1 h = 0.05 “Exact" 2.0 0.732639212 0.732638663 0.732638630 0.732638628 | Elementary Differential Equations with Boundary Value Problems_Page_731_Chunk3133 |
722 Answers to Selected Exercises 3.3.19 (p. 126) Runge-Kutta method x h = 0.0500 h = 0.0250 h = 0.0125 “Exact" 1.50 2.244025683 2.244024088 2.244023989 2.244023982 Runge-Kutta semilinear method x h = 0.0500 h = 0.0250 h = 0.0125 “Exact" 1.50 2.244025081 2.244024051 2.244023987 2.244023982 3.3.20 (p. 126) Runge-Kutta method x h = 0.1 h = 0.05 h = 0.025 “Exact" 1.0 0.056426886 0.056416137 0.056415552 0.056415515 Runge-Kutta semilinear method x h = 0.1 h = 0.05 h = 0.025 “Exact" 1.0 0.056415185 0.056415495 0.056415514 0.056415515 3.3.21 (p. 126) Runge-Kutta method x h = 0.1 h = 0.05 h = 0.025 “Exact" 1.0 54.695901186 54.727111858 54.729426250 54.729594761 Runge-Kutta semilinear method x h = 0.1 h = 0.05 h = 0.025 “Exact" 1.0 54.729099966 54.729561720 54.729592658 54.729594761 3.3.22 (p. 126) Runge-Kutta method x h = 0.1 h = 0.05 h = 0.025 “Exact" 3.0 1.361384082 1.361383812 1.361383809 1.361383810 Runge-Kutta semilinear method x h = 0.1 h = 0.05 h = 0.025 “Exact" 3.0 1.361456502 1.361388196 1.361384079 1.361383810 3.3.24 (p. 127) x h = .1 h = .05 h = .025 Exact 2.00 -1.000000000 -1.000000000 -1.000000000 -1.000000000 3.3.25 (p. 127) x h = .1 h = .05 h = .025 “Exact" 1.00 1.000000000 1.000000000 1.000000000 1.000000000 3.3.26 (p. 127) x h = .1 h = .05 h = .025 Exact 1.50 4.142171279 4.142170553 4.142170508 4.142170505 3.3.27 (p. 127) x h = .1 h = .05 h = .025 Exact 3.0 16.666666988 16.666666687 16.666666668 16.666666667 Section 4.1 Answers, pp. 138–140 4.1.1 (p. 138) Q = 20e−(t ln 2)/3200 g 4.1.2 (p. 138) 2 ln 10 ln 2 days 4.1.3 (p. 138) τ = 10 ln 2 ln 4/3 minutes | Elementary Differential Equations with Boundary Value Problems_Page_732_Chunk3134 |
Answers to Selected Exercises 723 4.1.4 (p. 138) τ ln(p0/p1) ln 2 4.1.5 (p. 138) tp tq = ln p ln q 4.1.6 (p. 138) k = 1 t2 −t1 ln Q1 Q2 4.1.7 (p. 138) 20 g 4.1.8 (p. 138) 50 ln 2 3 yrs 4.1.9 (p. 138) 25 2 ln 2% 4.1.10 (p. 138) (a) = 20 ln 3 yr (b). Q0 = 100000e−.5 4.1.11 (p. 138) (a) Q(t) = 5000 −4750e−t/10 (b) 5000 lbs 4.1.12 (p. 138) 1 25 yrs; 4.1.13 (p. 138) V = V0et ln 10/2 4 hours 4.1.14 (p. 138) 1500 ln 4 3 ln 2 yrs; 2−4/3Q0 4.1.15 (p. 138) W(t) = 20 −19e−t/20; limt→∞W(t) = 20 ounces 4.1.16 (p. 138) S(t) = 10(1 + e−t/10); limt→∞S(t) = 10 g 4.1.17 (p. 139) 10 gallons 4.1.18 (p. 139) V (t) = 15000 + 10000et/20 4.1.19 (p. 139) W(t) = 4 × 106(t + 1)2 dollars t years from now 4.1.20 (p. 139) p = 100 25 −24e−t/2 4.1.21 (p. 139) (a) P(t) = 1000e.06t + 50 e.06t −1 e.06/52 −1 (b) 5.64 × 10−4 4.1.22 (p. 139) (a) P ′ = rP −12M (b) P = 12M r (1 −ert) + P0ert (c) M ≈ rP0 12(1 −e−rN) (d) For (i) approximate M = $402.25, exact M = $402.80 for (ii) approximate M = $1206.05, exact M = $1206.93. 4.1.23 (p. 139) (a) T(α) = −1 r ln | Elementary Differential Equations with Boundary Value Problems_Page_733_Chunk3135 |
724 Answers to Selected Exercises 4.2.20 (p. 150) (a) cn = c 1 −e−rt/W n−1 X j=0 1 j! rt W j! (b) c (c) 0 4.2.21 (p. 150) Let c∞= c1W1 + c2W2 W1 + W2 , α = c2W 2 2 −c1W 2 1 W1 + W2 , and β = W1 + W2 W1W2 . Then: (a) c1(t) = c∞+ α W1 e−rβt, c2(t) = c∞−α W2 e−rβt (b) limt→∞c1(t) = limt→∞c2(t) = c∞ Section 4.3 Answers, pp. 160–162 4.3.1 (p. 160) v = −384 5 | Elementary Differential Equations with Boundary Value Problems_Page_734_Chunk3136 |
Answers to Selected Exercises 725 30v2 + y2(12y3 −15y2 −80y + 120) = 37 4.4.9 (p. 176) No equilibria if a < 0; 0 is unstable if a = 0; √a is stable and −√a is unstable if a > 0. * 4.4.10 (p. 176) 0 is a stable equilibrium if a ≤0; −√a and √a are stable and 0 is unstable if a > 0. 4.4.11 (p. 176) 0 is unstable if a ≤0; −√a and √a are unstable and 0 is stable if a > 0. 4.4.12 (p. 176) 0 is stable if a ≤0; 0 is stable and −√a and √a are unstable if a ≤0. 4.4.22 (p. 178) An equilibrium solution y of y′′ + p(y) = 0 is unstable if there’s an ϵ > 0 such that, for every δ > 0, there’s a solution of (A) with p (y(0) −y)2 + v2(0) < δ, but p (y(t) −y)2 + v2(t) ≥ ϵ for some t > 0. Section 4.5 Answers, pp. 190–192 4.5.1 (p. 190) y′ = − 2xy x2 + 3y2 4.5.2 (p. 190) y′ = − y2 (xy −1) 4.5.3 (p. 190) y′ = −y(x2 + y2 −2x2 ln |xy|) x(x2 + y2 −2y2 ln |xy|). 4.5.4 (p. 190) xy′ −y = −x1/2 2 4.5.5 (p. 190) y′ + 2xy = 4xex2 4.5.6 (p. 190) xy′ + y = 4x3 4.5.7 (p. 190) y′ −y = cos x −sin x 4.5.8 (p. 190) (1 + x2)y′ −2xy = (1 −x)2ex 4.5.10 (p. 190) y′g −yg′ = f ′g −fg′. 4.5.11 (p. 190) (x −x0)y′ = y −y0 4.5.12 (p. 190) y′(y2 −x2 + 1) + 2xy = 0 4.5.13 (p. 190) 2x(y −1)y′ −y2 + x2 + 2y = 0 4.5.14 (p. 190) (a) y = −81 + 18x, (9, 81) y = −1 + 2x, (1, 1) (b) y = −121 + 22x, (11, 121) y = −1 + 2x, (1, 1) (c) y = −100 −20x, (−10, 100) y = −4 −4x, (−2, 4) (d) y = −25 −10x, (−5, 25) y = −1 −2x, (−1, 1) 4.5.15 (p. 190) (e) y = 5 + 3x 4 , (−3/5, 4/5) y = −5 −4x 3 , (4/5, −3/5) 4.5.17 (p. 191) (a) y = −1 2(1 + x), (1, −1); y = 5 2 + x 10, (25, 5) (b) y = 1 4(4 + x), (4, 2) y = −1 4(4 + x), (4, −2); (c) y = 1 2 (1 + x),(1, 1) y = 7 2 + x 14, (49, 7) (d) y = −1 2(1 + x),(1, −1) y = −5 2 −x 10 , (25, −5) 4.5.18 (p. 191) y = 2x2 4.5.19 (p. 192) y = cx p |x2 −1| 4.5.20 (p. 192) y = y1 + c(x −x1) 4.5.21 (p. 192) y = −x3 2 −x 2 4.5.22 (p. 192) y = −x ln |x| + cx 4.5.23 (p. 192) y = √ 2x + 4 4.5.24 (p. 192) y = p x2 −3 4.5.25 (p. 192) y = kx2 4.5.26 (p. 192) (y −x)3(y + x) = k 4.5.27 (p. 192) y2 = −x + k 4.5.28 (p. 192) y2 = −1 2 ln(1 + 2x2) + k 4.5.29 (p. 192) y2 = −2x −ln(x −1)2 + k 4.5.30 (p. 192) y = 1 + r 9 −x2 2 ; those with c > 0 4.5.33 (p. 192) tan−1 y x −1 2 ln(x2 + y2) = k 4.5.34 (p. 192) 1 2 ln(x2 + y2) + (tan α) tan−1 y x = k Section 5.1 Answers, pp. 203–210 5.1.1 (p. 203) (c) y = −2e2x + e5x (d) y = (5k0 −k1)e2x 3 + (k1 −2k0)e5x 3 . 5.1.2 (p. 203) (c) y = ex(3 cos x −5 sin x) (d) y = ex (k0 cos x + (k1 −k0) sin x) 5.1.3 (p. 204) (c) y = ex(7 −3x) (d) y = ex (k0 + (k1 −k0)x) | Elementary Differential Equations with Boundary Value Problems_Page_735_Chunk3137 |
726 Answers to Selected Exercises 5.1.4 (p. 204) (a) y = c1 x −1 + c2 x + 1 (b) y = 2 x −1 − 3 x + 1 ; (−1, 1) 5.1.5 (p. 204) (a) ex (b) e2x cos x (c) x2 + 2x −2 (d) −5 6x−5/6 (e) −1 x2 (f) (x ln |x|)2 (g) e2x 2√x 5.1.6 (p. 204) 0 5.1.7 (p. 204) W(x) = (1 −x2)−1 5.1.8 (p. 205) W(x) = 1 x 5.1.10 (p. 205) y2 = e−x 5.1.11 (p. 205) y2 = xe3x 5.1.12 (p. 205) y2 = xeax 5.1.13 (p. 205) y2 = 1 x 5.1.14 (p. 205) y2 = x ln x 5.1.15 (p. 205) y2 = xa ln x 5.1.16 (p. 205) y2 = x1/2e−2x 5.1.17 (p. 205) y2 = x 5.1.18 (p. 205) y2 = x sin x 5.1.19 (p. 205) y2 = x1/2 cos x 5.1.20 (p. 205) y2 = xe−x 5.1.21 (p. 205) y2 = 1 x2 −4 5.1.22 (p. 205) y2 = e2x 5.1.23 (p. 205) y2 = x2 5.1.35 (p. 207) (a) y′′ −2y′ + 5y = 0 (b) (2x −1)y′′ −4xy′ + 4y = 0 (c) x2y′′ −xy′ + y = 0 (d) x2y′′ + xy′ + y = 0 (e) y′′ −y = 0 (f) xy′′ −y′ = 0 5.1.37 (p. 207) (c) y = k0y1 + k1y2 5.1.38 (p. 208) y1 = 1, y2 = x −x0; y = k0 + k1(x −x0) 5.1.39 (p. 208) y1 = cosh(x −x0), y2 = sinh(x −x0); y = k0 cosh(x −x0) + k1 sinh(x −x0) 5.1.40 (p. 208) y1 = cos ω(x −x0), y2 = 1 ω sin ω(x −x0) y = k0 cos ω(x −x0) + k1 ω sin ω(x −x0) 5.1.41 (p. 208) y1 = 1 1 −x2 , y2 = x 1 −x2 y = k0 + k1x 1 −x2 5.1.42 (p. 209) (c) k0 = k1 = 0; y = c1x2 + c2x3, x ≥0, c1x2 + c3x3, x < 0 (d) (0, ∞) if x0 > 0, (−∞, 0) if x0 < 0 5.1.43 (p. 209) (c) k0 = 0, k1 arbitrary y = k1x + c2x2 5.1.44 (p. 210) (c) k0 = k1 = 0 y = a1x3 + a2x4, x ≥0, b1x3 + b2x4, x < 0 (d) (0, ∞) if x0 > 0, (−∞, 0) if x0 < 0 Section 5.2 Answers, pp. 217–220 5.2.1 (p. 217) y = c1e−6x+c2ex 5.2.2 (p. 217) y = e2x(c1 cos x+c2 sin x) 5.2.3 (p. 217) y = c1e−7x+c2e−x 5.2.4 (p. 217) y = e2x(c1 + c2x) 5.2.5 (p. 217) y = e−x(c1 cos 3x + c2 sin 3x) 5.2.6 (p. 217) y = e−3x(c1 cos x + c2 sin x) 5.2.7 (p. 217) y = e4x(c1 + c2x) 5.2.8 (p. 217) y = c1 + c2e−x 5.2.9 (p. 217) y = ex(c1 cos √ 2x + c2 sin √ 2x) 5.2.10 (p. 217) y = e−3x(c1 cos 2x + c2 sin 2x) 5.2.11 (p. 217) y = e−x/2 c1 cos 3x 2 + c2 sin 3x 2 5.2.12 (p. 217) y = c1e−x/5 + c2ex/2 5.2.13 (p. 218) y = e−7x(2 cos x −3 sin x) 5.2.14 (p. 218) y = 4ex/2 + 6e−x/3 5.2.15 (p. 218) y = 3ex/3 −4e−x/2 5.2.16 (p. 218) y = e−x/2 3 + 3e3x/2 4 5.2.17 (p. 218) y = e3x/2(3 −2x) 5.2.18 (p. 218) y = 3e−4x −4e−3x 5.2.19 (p. 218) y = 2xe3x 5.2.20 (p. 218) y = ex/6(3+2x) 5.2.21 (p. 218) y = e−2x 3 cos √ 6x + 2 √ 6 3 sin √ 6x 5.2.23 (p. 218) y = 2e−(x−1) −3e−2(x−1) 5.2.24 (p. 219) y = 1 3e−(x−2) −2 3 e7(x−2) 5.2.25 (p. 219) y = e7(x−1) (2 −3(x −1)) 5.2.26 (p. 219) y = e−(x−2)/3 (2 −4(x −2)) 5.2.27 (p. 219) y = 2 cos 2 3 x −π 4 −3 sin 2 3 x −π 4 5.2.28 (p. 219) y = 2 cos √ 3 x −π 3 − 1 √ 3 sin √ 3 | Elementary Differential Equations with Boundary Value Problems_Page_736_Chunk3138 |
x −π 3 5.2.30 (p. 219) y = k0 r2 −r1 | Elementary Differential Equations with Boundary Value Problems_Page_736_Chunk3139 |
Answers to Selected Exercises 727 5.2.32 (p. 219) y = eλ(x−x0) h k0 cos ω(x −x0) + k1 −λk0 ω sin ω(x −x0) i Section 5.3 Answers, pp. 227–229 5.3.1 (p. 227) yp = −1 + 2x + 3x2; y = −1 + 2x + 3x2 + c1e−6x + c2ex 5.3.2 (p. 227) yp = 1 + x; y = 1 + x + e2x(c1 cos x + c2 sin x) 5.3.3 (p. 227) yp = −x + x3; y = −x + x3 + c1e−7x + c2e−x 5.3.4 (p. 227) yp = 1 −x2; y = 1 −x2 + e2x(c1 + c2x) 5.3.5 (p. 227) yp = 2x + x3; y = 2x + x3 + e−x(c1 cos 3x + c2 sin 3x); y = 2x + x3 + e−x(2 cos 3x + 3 sin 3x) 5.3.6 (p. 227) yp = 1 + 2x; y = 1 + 2x + e−3x(c1 cos x + c2 sin x); y = 1 + 2x + e−3x(cos x −sin x) 5.3.8 (p. 227) yp = 2 x 5.3.9 (p. 227) yp = 4x1/2 5.3.10 (p. 227) yp = x3 2 5.3.11 (p. 227) yp = 1 x3 5.3.12 (p. 227) yp = 9x1/3 5.3.13 (p. 227) yp = 2x4 13 5.3.16 (p. 228) yp = e3x 3 ; y = e3x 3 + c1e−6x + c2ex 5.3.17 (p. 228) yp = e2x; y = e2x(1 + c1 cos x + c2 sin x) 5.3.18 (p. 228) y = −2e−2x; y = −2e−2x + c1e−7x + c2e−x; y = −2e−2x −e−7x + e−x 5.3.19 (p. 228) yp = ex; y = ex + e2x(c1 + c2x); y = ex + e2x(1 −3x) 5.3.20 (p. 228) yp = 4 45ex/2; y = 4 45ex/2 + e−x(c1 cos 3x + c2 sin 3x) 5.3.21 (p. 228) yp = e−3x; y = e−3x(1 + c1 cos x + c2 sin x) 5.3.24 (p. 228) yp = cos x −sin x; y = cos x −sin x + e4x(c1 + c2x) 5.3.25 (p. 228) yp = cos 2x −2 sin 2x; y = cos 2x −2 sin 2x + c1 + c2e−x 5.3.26 (p. 228) yp = cos 3x; y = cos 3x + ex(c1 cos √ 2x + c2 sin √ 2x) 5.3.27 (p. 228) yp = cos x + sin x; y = cos x + sin x + e−3x(c1 cos 2x + c2 sin 2x) 5.3.28 (p. 228) yp = −2 cos 2x + sin 2x; y = −2 cos 2x + sin 2x + c1e−4x + c2e−3x y = −2 cos 2x + sin 2x + 2e−4x −3e−3x 5.3.29 (p. 228) yp = cos 3x −sin 3x; y = cos 3x −sin 3x + e3x(c1 + c2x) y = cos 3x −sin 3x + e3x(1 + 2x) 5.3.30 (p. 228) y = 1 ω2 0 −ω2 (M cos ωx + N sin ωx) + c1 cos ω0x + c2 sin ω0x 5.3.33 (p. 229) yp = −1 + 2x + 3x2 + e3x 3 ; y = −1 + 2x + 3x2 + e3x 3 + c1e−6x + c2ex 5.3.34 (p. 229) yp = 1 + x + e2x; y = 1 + x + e2x(1 + c1 cos x + c2 sin x) 5.3.35 (p. 229) yp = −x + x3 −2e−2x; y = −x + x3 −2e−2x + c1e−7x + c2e−x 5.3.36 (p. 229) yp = 1 −x2 + ex; y = 1 −x2 + ex + e2x(c1 + c2x) 5.3.37 (p. 229) yp = 2x + x3 + 4 45ex/2; y = 2x + x3 + 4 45 ex/2 + e−x(c1 cos 3x + c2 sin 3x) 5.3.38 (p. 229) yp = 1 + 2x + e−3x; y = 1 + 2x + e−3x(1 + c1 cos x + c2 sin x) Section 5.4 Answers, pp. 235–238 5.4.1 (p. 235) yp = e3x −1 4 + x 2 5.4.2 (p. 235) yp = e−3x 1 −x 4 5.4.3 (p. 235) yp = ex 2 −3x 4 5.4.4 (p. 235) yp = e2x(1−3x+x2) 5.4.5 (p. 235) yp | Elementary Differential Equations with Boundary Value Problems_Page_737_Chunk3140 |
= e−x(1+x2) 5.4.6 (p. 235) yp = ex(−2+x+2x2) 5.4.7 (p. 235) yp = xe−x 1 6 + x 2 5.4.8 (p. 235) yp = xex(1 + 2x) 5.4.9 (p. 235) yp = xe3x −1 + x 2 5.4.10 (p. 235) yp = xe2x(−2+x) 5.4.11 (p. 235) yp = x2e−x 1 + x 2 5.4.12 (p. 235) yp = x2ex 1 2 −x 5.4.13 (p. 235) yp = x2e2x 2 (1 −x + x2) 5.4.14 (p. 235) yp = x2e−x/3 27 (3 −2x + x2) | Elementary Differential Equations with Boundary Value Problems_Page_737_Chunk3141 |
728 Answers to Selected Exercises 5.4.15 (p. 235) y = e3x 4 (−1 + 2x) + c1ex + c2e2x 5.4.16 (p. 235) y = ex(1 −2x) + c1e2x + c2e4x 5.4.17 (p. 235) y = e2x 5 (1 −x) + e−3x(c1 + c2x) 5.4.18 (p. 235) y = xex(1 −2x) + c1ex + c2e−3x 5.4.19 (p. 235) y = ex x2(1 −2x) + c1 + c2x 5.4.20 (p. 236) y = −e2x(1 + x) + 2e−x −e5x 5.4.21 (p. 236) y = xe2x + 3ex −e−4x 5.4.22 (p. 236) y = e−x(2 + x −2x2) −e−3x 5.4.23 (p. 236) y = e−2x(3 −x) −2e5x 5.4.24 (p. 236) yp = −ex 3 (1 −x) + e−x(3 + 2x) 5.4.25 (p. 236) yp = ex(3 + 7x) + xe3x 5.4.26 (p. 236) yp = x3e4x + 1 + 2x + x2 5.4.27 (p. 236) yp = xe2x(1 −2x) + xex 5.4.28 (p. 236) yp = ex(1 + x) + x2e−x 5.4.29 (p. 236) yp = x2e−x + e3x(1 −x2) 5.4.31 (p. 237) yp = 2e2x 5.4.32 (p. 237) yp = 5xe4x 5.4.33 (p. 237) yp = x2e4x 5.4.34 (p. 237) yp = −e3x 4 (1+2x−2x2) 5.4.35 (p. 237) yp = xe3x(4−x+2x2) 5.4.36 (p. 237) yp = x2e−x/2(−1 + 2x + 3x2) 5.4.37 (p. 237) (a) y = e−x 4 3 x3/2 + c1x + c2 (b) y = e−3x x2 4 (2 ln x −3) + c1x + c2 (c) y = e2x [(x + 1) ln |x + 1| + c1x + c2] (d) y = e−x/2 x ln |x| + x3 6 + c1x + c2 5.4.39 (p. 238) (a) ex(3 + x) + c (b) −e−x(1 + x)2 + c (c) −e−2x 8 (3 + 6x + 6x2 + 4x3) + c (d) ex(1 + x2) + c (e) e3x(−6 + 4x + 9x2) + c (f) −e−x(1 −2x3 + 3x4) + c 5.4.40 (p. 238) (−1)kk!eαx αk+1 k X r=0 (−αx)r r! + c Section 5.5 Answers, pp. 244–248 5.5.1 (p. 244) yp = cos x + 2 sin x 5.5.2 (p. 244) yp = cos x + (2 −2x) sin x 5.5.3 (p. 244) yp = ex(−2 cos x + 3 sin x) 5.5.4 (p. 244) yp = e2x 2 (cos 2x −sin 2x) 5.5.5 (p. 244) yp = −ex(x cos x −sin x) 5.5.6 (p. 244) yp = e−2x(1 −2x)(cos 3x −sin 3x) 5.5.7 (p. 245) yp = x(cos 2x −3 sin 2x) 5.5.8 (p. 245) yp = −x [(2 −x) cos x + (3 −2x) sin x] 5.5.9 (p. 245) yp = x h x cos x 2 −3 sin x 2 i 5.5.10 (p. 245) yp = xe−x(3 cos x + 4 sin x) 5.5.11 (p. 245) yp = xex [(−1 + x) cos 2x + (1 + x) sin 2x] 5.5.12 (p. 245) yp = −(14 −10x) cos x −(2 + 8x −4x2) sin x. 5.5.13 (p. 245) yp = (1 + 2x + x2) cos x + (1 + 3x2) sin x 5.5.14 (p. 245) yp = x2 2 (cos 2x −sin 2x) 5.5.15 (p. 245) yp = ex(x2 cos x + 2 sin x) 5.5.16 (p. 245) yp = ex(1 −x2)(cos x + sin x) 5.5.17 (p. 245) yp = ex(x2 −x3)(cos x + sin x) 5.5.18 (p. 245) yp = e−x [(1 + 2x) cos x −(1 −3x) sin x] 5.5.19 (p. 245) yp = x(2 cos 3x −sin 3x) 5.5.20 (p. 245) yp = −x3 cos x + (x + 2x2) sin x 5.5.21 (p. 245) yp = −e−x (x + x2) cos x −(1 + 2x) sin x 5.5.22 (p. 245) y = ex(2 cos x + 3 sin x) + 3ex −e6x 5.5.23 (p. 245) y = ex [(1 + 2x) cos x + (1 −3x) sin x] 5.5.24 (p. 245) y = ex(cos x−2 sin x)+e−3x(cos x+sin x) 5.5.25 (p. 245) y = e3x [(2 + 2x) cos x −(1 + 3x) sin x] 5.5.26 (p. 245) y = e3x [(2 + 3x) cos x + (4 −x) sin x]+3ex−5e2x 5.5.27 | Elementary Differential Equations with Boundary Value Problems_Page_738_Chunk3142 |
(p. 245) yp = xe3x −ex 5 (cos x −2 sin x) 5.5.28 (p. 245) yp = x(cos x + 2 sin x) −ex 2 (1 −x) + e−x 2 5.5.29 (p. 245) yp = −xex 2 (2 + x) + 2xe2x + 1 10(3 cos x + sin x) | Elementary Differential Equations with Boundary Value Problems_Page_738_Chunk3143 |
Answers to Selected Exercises 729 5.5.30 (p. 245) yp = xex(cos x + x sin x) + e−x 25 (4 + 5x) + 1 + x + x2 2 5.5.31 (p. 245) yp = x2e2x 6 (3 + x) −e2x(cos x −sin x) + 3e3x + 1 4(2 + x) 5.5.32 (p. 245) y = (1 −2x + 3x2)e2x + 4 cos x + 3 sin x 5.5.33 (p. 245) y = xe−2x cos x + 3 cos 2x 5.5.34 (p. 245) y = −3 8 cos 2x + 1 4 sin 2x + e−x −13 8 e−2x −3 4xe−2x 5.5.40 (p. 248) (a) 2x cos x −(2 −x2) sin x + c (b) −ex 2 (1 −x2) cos x −(1 −x)2 sin x + c (c) −e−x 25 [(4 + 10x) cos 2x −(3 −5x) sin 2x] + c (d) −e−x 2 (1 + x)2 cos x −(1 −x2) sin x + c (e) −ex 2 x(3 −3x + x2) cos x −(3 −3x + x3) sin x + c (f) −ex [(1 −2x) cos x + (1 + x) sin x] + c (g) e−x [x cos x + x(1 + x) sin x] + c Section 5.6 Answers, pp. 253–255 5.6.1 (p. 253) y = 1 −2x + c1e−x + c2xex; {e−x, xex} 5.6.2 (p. 253) y = 4 3x2 + c1x + c2 x ; {x, 1/x} 5.6.3 (p. 253) y = x(ln |x|)2 2 + c1x + c2x ln |x|; {x, x ln |x|} 5.6.4 (p. 253) y = (e2x + ex) ln(1 + e−x) + c1e2x + c2ex; {e2x, ex} 5.6.5 (p. 253) y = ex4 5x7/2 + c1 + c2x ; {ex, xex} 5.6.6 (p. 253) y = ex(2x3/2 + x1/2 ln x + c1x1/2 + c2x−1/2); {x1/2ex, x−1/2e−x} 5.6.7 (p. 253) y = ex(x sin x + cos x ln | cos x| + c1 cos x + c2 sin x); {ex cos x, ex sin x} 5.6.8 (p. 253) y = e−x2(2e−2x + c1 + c2x); {e−x2, xe−x2} 5.6.9 (p. 253) y = 2x + 1 + c1x2 + c2 x2 ; {x2, 1/x2} 5.6.10 (p. 253) y = xe2x 9 + xe−x(c1 + c2x); {xe−x, x2e−x} 5.6.11 (p. 253) y = xexx 3 + c1 + c2 x2 ; {xex, ex/x} 5.6.12 (p. 253) y = −(2x −1)2ex 8 + c1ex + c2xe−x; {ex, xe−x} 5.6.13 (p. 253) y = x4 + c1x2 + c2x2 ln |x|; {x2, x2 ln |x|} 5.6.14 (p. 253) y = e−x(x3/2 + c1 + c2x1/2); {e−x, x1/2e−x} 5.6.15 (p. 253) y = ex(x+c1+c2x2); {ex, x2ex} 5.6.16 (p. 253) y = x1/2 e2x 2 + c1 + c2ex ; {x1/2, x1/2ex} 5.6.17 (p. 253) y = −2x2 ln x + c1x2 + c2x4; {x2, x4} 5.6.18 (p. 253) {ex, ex/x} 5.6.19 (p. 253) {x2, x3} 5.6.20 (p. 253) {ln |x|,x ln |x|} 5.6.21 (p. 253) {sin √x, cos √x} 5.6.22 (p. 253) {ex, x3ex} 5.6.23 (p. 253) {xa, xa ln x} 5.6.24 (p. 253) {x sin x, x cos x} 5.6.25 (p. 253) {e2x, x2e2x} 5.6.26 (p. 253) {x1/2, x1/2 cos x} 5.6.27 (p. 253) {x1/2e2x, x1/2e−2x} 5.6.28 (p. 253) {1/x, e2x} 5.6.29 (p. 253) {ex, x2} 5.6.30 (p. 253) {e2x, x2e2x} 5.6.31 (p. 253) y = x4 + 6x2 −8x2 ln |x| 5.6.32 (p. 253) y = 2e2x −xe−x 5.6.33 (p. 254) y = (x + 1) 4 −ex(3 −2x) + 7e−x 5.6.34 (p. 254) y = x2 4 + x 5.6.35 (p. 254) y = (x + 2)2 6(x −2) + 2x x2 −4 | Elementary Differential Equations with Boundary Value Problems_Page_739_Chunk3144 |
730 Answers to Selected Exercises 5.6.38 (p. 254) (a) y = −kc1 sin kx + kc2 cos kx c1 cos kx + c2 sin kx (b) y = c1 + 2c2ex c1 + c2ex (c) y = −6c1 + c2e7x c1 + c2e7x (d) y = −7c1 + c2e6x c1 + c2e6x (e) y = −(7c1 −c2) cos x + (c1 + 7c2) sin x c1 cos x + c2 sin x (f) y = −2c1 + 3c2e5x/6 6(c1 + c2e5x/6) (g) y = c1 + c2(x + 6) 6(c1 + c2x) 5.6.39 (p. 254) (a) y = c1 + c2ex(1 + x) x(c1 + c2ex) (b) y = −2c1x + c2(1 −2x2) c1 + c2x (c) y = −c1 + c2e2x(x + 1) c1 + c2xe2x (d) y = 2c1 + c2e−3x(1 −x) c1 + c2xe−3x (e) y = (2c2x −c1) cos x −(2c1x + c2) sin x 2x(c1 cos x + c2 sin x) (f) y = c1 + 7c2x6 x(c1 + c2x6) Section 5.7 Answers, pp. 262–264 5.7.1 (p. 262) yp = −cos 3x ln | sec 3x + tan 3x| 9 5.7.2 (p. 262) yp = −sin 2x ln | cos 2x| 4 + x cos 2x 2 5.7.3 (p. 262) yp = 4ex(1 + ex) ln(1 + e−x) 5.7.4 (p. 262) yp = 3ex(cos x ln | cos x| + x sin x) 5.7.5 (p. 262) yp = 8 5x7/2ex 5.7.6 (p. 262) yp = ex ln(1 −e−2x) −e−x ln(e2x −1) 5.7.7 (p. 263) yp = 2(x2 −3) 3 5.7.8 (p. 263) yp = e2x x 5.7.9 (p. 263) yp = x1/2ex ln x 5.7.10 (p. 263) yp = e−x(x+2) 5.7.11 (p. 263) yp = −4x5/2 5.7.12 (p. 263) yp = −2x2 sin x−2x cos x 5.7.13 (p. 263) yp = −xe−x(x + 1) 2 5.7.14 (p. 263) yp = − √x cos √x 2 5.7.15 (p. 263) yp = 3x4ex 2 5.7.16 (p. 263) yp = xa+1 5.7.17 (p. 263) yp = x2 sin x 2 5.7.18 (p. 263) yp = −2x2 5.7.19 (p. 263) yp = −e−x sin x 5.7.20 (p. 263) yp = − √x 2 5.7.21 (p. 263) yp = x3/2 4 5.7.22 (p. 263) yp = −3x2 5.7.23 (p. 263) yp = x3ex 2 5.7.24 (p. 263) yp = −4x3/2 15 5.7.25 (p. 263) yp = x3ex 5.7.26 (p. 263) yp = xex 5.7.27 (p. 263) yp = x2 5.7.28 (p. 263) yp = xex(x −2) 5.7.29 (p. 263) yp = √xex(x −1)/4 5.7.30 (p. 263) y = e2x(3x2 −2x + 6) 6 + xe−x 3 5.7.31 (p. 263) y = (x −1)2 ln(1 −x) + 2x2 −5x + 3 5.7.32 (p. 263) y = (x2−1)ex−5(x−1) 5.7.33 (p. 264) y = x(x2 + 6) 3(x2 −1) 5.7.34 (p. 264) y = −x2 2 + x + 1 2x2 5.7.35 (p. 264) y = x2(4x + 9) 6(x + 1) 5.7.38 (p. 264) (a) y = k0 cosh x + k1 sinh x + Z x 0 sinh(x −t)f(t) dt (b) y′ = k0 sinh x + k1 cosh x + Z x 0 cosh(x −t)f(t)dt 5.7.39 (p. 264) (a) y(x) = k0 cos x + k1 sin x + Z x 0 sin(x −t)f(t)dt (b) y′(x) = −k0 sin x + k1 cos x + R x 0 cos(x −t)f(t) dt | Elementary Differential Equations with Boundary Value Problems_Page_740_Chunk3145 |
Answers to Selected Exercises 731 Section 6.1 Answers, pp. 277–278 6.1.1 (p. 277) y = 3 cos 4 √ 6t − 1 2 √ 6 sin 4 √ 6t ft 6.1.2 (p. 277) y = −1 4 cos 8 √ 5t − 1 4 √ 5 sin 8 √ 5t ft 6.1.3 (p. 277) y = 1.5 cos 14 √ 10t cm 6.1.4 (p. 277) y = 1 4 cos 8t −1 16 sin 8t ft; R = √ 17 16 ft; ω0 = 8 rad/s; T = π/4 s; φ ≈−.245 rad ≈−14.04◦; 6.1.5 (p. 277) y = 10 cos 14t + 25 14 sin 14t cm; R = 5 14 √ 809 cm; ω0 = 14 rad/s; T = π/7 s; φ ≈.177 rad ≈10.12◦ 6.1.6 (p. 277) y = −1 4 cos √ 70 t + 2 √ 70 sin √ 70 t m; R = 1 4 r 67 35 m ω0 = √ 70 rad/s; T = 2π/ √ 70 s; φ ≈2.38 rad ≈136.28◦ 6.1.7 (p. 277) y = 2 3 cos 16t −1 4 sin 16t ft 6.1.8 (p. 278) y = 1 2 cos 8t −3 8 sin 8t ft 6.1.9 (p. 278) .72 m 6.1.10 (p. 278) y = 1 3 sin t + 1 2 cos 2t + 5 6 sin 2t ft 6.1.11 (p. 278) y = 16 5 4 sin t 4 −sin t 6.1.12 (p. 278) y = −1 16 sin 8t + 1 3 cos 4 √ 2t − 1 8 √ 2 sin 4 √ 2t 6.1.13 (p. 278) y = −t cos 8t −1 6 cos 8t + 1 8 sin 8t ft 6.1.14 (p. 278) T = 4 √ 2 s 6.1.15 (p. 278) ω = 8 rad/s y = −t 16(−cos 8t + 2 sin 8t) + 1 128 sin 8t ft 6.1.16 (p. 278) ω = 4 √ 6 rad/s; y = −t √ 6 h8 3 cos 4 √ 6t + 4 sin 4 √ 6t i + 1 9 sin 4 √ 6t ft 6.1.17 (p. 278) y = t 2 cos 2t −t 4 sin 2t + 3 cos 2t + 2 sin 2t m 6.1.18 (p. 278) y = y0 cos ω0t + v0 ω0 sin ω0t; R = 1 ω0 p (ω0y0)2 + (v0)2; cos φ = y0ω0 p (ω0y0)2 + (v0)2 ; sin φ = v0 p (ω0y0)2 + (v0)2 6.1.19 (p. 278) The object with the longer period weighs four times as much as the other. 6.1.20 (p. 278) T2 = √ 2T1, where T1 is the period of the smaller object. 6.1.21 (p. 278) k1 = 9k2, where k1 is the spring constant of the system with the shorter period. Section 6.2 Answers, pp. 287–289 6.2.1 (p. 287) y = e−2t 2 (3 cos 2t −sin 2t) ft; r 5 2e−2t ft 6.2.2 (p. 287) y = −e−t 3 cos 3t + 1 3 sin 3t ft √ 82 3 e−t ft 6.2.3 (p. 287) y = e−16t 1 4 + 10t ft 6.2.4 (p. 287) y = −e−3t 4 (5 cos t + 63 sin t) ft 6.2.5 (p. 287) 0 ≤c < 8 lb-sec/ft 6.2.6 (p. 287) y = 1 2e−3t cos √ 91t + 11 √ 91 sin √ 91t ft 6.2.7 (p. 287) y = −e−4t 3 (2 + 8t) ft 6.2.8 (p. 287) y = e−10t 9 cos 4 √ 6t + 45 2 √ 6 sin 4 √ 6t cm 6.2.9 (p. 287) y = e−3t/2 3 2 cos √ 41 2 t + 9 2 √ 41 sin √ 41 2 t ft | Elementary Differential Equations with Boundary Value Problems_Page_741_Chunk3146 |
732 Answers to Selected Exercises 6.2.10 (p. 287) y = e−3 2 t 1 2 cos √ 119 2 t − 9 2 √ 119 sin √ 119 2 t ft 6.2.11 (p. 287) y = e−8t 1 4 cos 8 √ 2t − 1 4 √ 2 sin 8 √ 2t ft 6.2.12 (p. 287) y = e−t −1 3 cos 3 √ 11t + 14 9 √ 11 sin 3 √ 11t ft 6.2.13 (p. 287) yp = 22 61 cos 2t + 2 61 sin 2t ft 6.2.14 (p. 288) y = −2 3(e−8t −2e−4t) 6.2.15 (p. 288) y = e−2t 1 10 cos 4t −1 5 sin 4t m 6.2.16 (p. 288) y = e−3t(10 cos t −70 sin t) cm 6.2.17 (p. 288) yp = −2 15 cos 3t + 1 15 sin 3t ft 6.2.18 (p. 288) yp = 11 100 cos 4t + 27 100 sin 4t cm 6.2.19 (p. 288) yp = 42 73 cos t + 39 73 sin t ft 6.2.20 (p. 288) y = −1 2 cos 2t + 1 4 sin 2t m 6.2.21 (p. 288) yp = 1 cω0 (−β cos ω0t + α sin ω0t) 6.2.24 (p. 288) y = e−ct/2m y0 cos ω1t + 1 ω1 (v0 + cy0 2m) sin ω1t 6.2.25 (p. 288) y = r2y0 −v0 r2 −r1 er1t + v0 −r1y0 r2 −r1 er2t 6.2.26 (p. 289) y = er1t(y0 + (v0 −r1y0)t) Section 6.3 Answers, pp. 294–295 6.3.1 (p. 294) I = e−15t 2 cos 5 √ 15t − 6 √ 31 sin 5 √ 31t 6.3.2 (p. 294) I = e−20t(2 cos 40t −101 sin 40t) 6.3.3 (p. 294) I = −200 3 e−10t sin 30t 6.3.4 (p. 294) I = −10e−30t(cos 40t + 18 sin 40t) 6.3.5 (p. 294) I = −e−40t(2 cos 30t −86 sin 30t) 6.3.6 (p. 294) Ip = −1 3(cos 10t + 2 sin 10t) 6.3.7 (p. 294) Ip = 20 37(cos 25t −6 sin 25t) 6.3.8 (p. 294) Ip = 3 13 (8 cos 50t −sin 50t) 6.3.9 (p. 294) Ip = 20 123 (17 sin 100t −11 cos 100t) 6.3.10 (p. 294) Ip = −45 52(cos 30t + 8 sin 30t) 6.3.12 (p. 295) ω0 = 1/ √ LC maximum amplitude = √ U 2 + V 2/R Section 6.4 Answers, pp. 301–302 6.4.1 (p. 301) If e = 1, then Y 2 = ρ(ρ −2X); if e ̸= 1 X + eρ 1 −e2 2 + Y 2 1 −e2 = ρ2 (1 −e2)2 if ; e < 1 let X0 = − eρ 1 −e2 , a = ρ 1 −e2 , b = ρ √ 1 −e2 . 6.4.2 (p. 302) Let h = r2 0θ′ 0; then ρ = h2 k , e = " ρ r0 −1 2 + ρr′ 0 h 2#1/2 . If e = 0, then θ0 is undefined, but also irrelevant if e ̸= 0 then φ = θ0 −α, where −π ≤α < π, cos α = 1 e ρ r0 −1 and sin α = ρr′ 0 eh . 6.4.3 (p. 302) (a) e = γ2 −γ1 γ1 + γ2 (b) r0 = Rγ1, r′ 0 = 0, θ0 arbitrary, θ′ 0 = 2gγ2 Rγ3 1(γ1 + γ2) 1/2 6.4.4 (p. 302) f(r) = −mh2 6c r4 + 1 r3 6.4.5 (p. 302) f(r) = −mh2(γ2 + 1) r3 6.4.6 (p. 302) (a) d2u dθ2 + 1 −k h2 u = 0, u(θ0) = 1 r0 , du(θ0) dθ = −r′ 0 h . (b) with γ = | Elementary Differential Equations with Boundary Value Problems_Page_742_Chunk3147 |
Answers to Selected Exercises 733 1 −k h2 1/2 : (i) r = r0 cosh γ(θ −θ0) −r0r′ 0 γh sinh γ(θ −θ0) −1 (ii) r = r0 1 −r0r′ 0 h (θ −θ0) −1 ; (iii) r = r0 cos γ(θ −θ0) −r0r′ 0 γh sin γ(θ −θ0) −1 Section 7.1 Answers, pp. 316–319 7.1.1 (p. 316) (a) R = 2; I = (−1, 3); (b) R = 1/2; I = (3/2, 5/2) (c) R = 0; (d) R = 16; I = (−14, 18) (e) R = ∞; I = (−∞, ∞) (f) R = 4/3; I = (−25/3, −17/3) 7.1.3 (p. 316) (a) R = 1; I = (0, 2) (b) R = √ 2; I = (−2 − √ 2, −2 + √ 2); (c) R = ∞; I = (−∞, ∞) (d) R = 0 (e) R = √ 3; I = (− √ 3, √ 3) (f) R = 1 I = (0, 2) 7.1.5 (p. 316) (a) R = 3; I = (0, 6) (b) R = 1; I = (−1, 1) (c) R = 1/ √ 3 I = (3 −1/ √ 3, 3 + 1/ √ 3) (d) R = ∞; I = (−∞, ∞) (e) R = 0 (f) R = 2; I = (−1, 3) 7.1.11 (p. 317) bn = 2(n + 2)(n + 1)an+2 + (n + 1)nan+1 + (n + 3)an 7.1.12 (p. 317) b0 = 2a2 −2a0 bn = (n + 2)(n + 1)an+2 + [3n(n −1) −2]an + 3(n −1)an−1, n ≥1 7.1.13 (p. 317) bn = (n + 2)(n + 1)an+2 + 2(n + 1)an+1 + (2n2 −5n + 4)an 7.1.14 (p. 317) bn = (n + 2)(n + 1)an+2 + 2(n + 1)an+1 + (n2 −2n + 3)an 7.1.15 (p. 318) bn = (n + 2)(n + 1)an+2 + (3n2 −5n + 4)an 7.1.16 (p. 318) b0 = −2a2 + 2a1 + a0, bn = −(n + 2)(n + 1)an+2 + (n + 1)(n + 2)an+1 + (2n + 1)an + an−1, n ≥2 7.1.17 (p. 318) b0 = 8a2 + 4a1 −6a0, bn = 4(n + 2)(n + 1)an+2 + 4(n + 1)2an+1 + (n2 + n −6)an −3an−1, n ≥1 7.1.21 (p. 319) b0 = (r + 1)(r + 2)a0, bn = (n + r + 1)(n + r + 2)an −(n + r −2)2an−1, n ≥1. 7.1.22 (p. 319) b0 = (r −2)(r + 2)a0, bn = (n + r −2)(n + r + 2)an + (n + r + 2)(n + r −3)an−1, n ≥14 7.1.23 (p. 319) b0 = (r −1)2a0, b1 = r2a1 + (r + 2)(r + 3)a0, bn = (n + r −1)2an + (n + r + 1)(n + r + 2)an−1 + (n + r −1)an−2, n ≥2 7.1.24 (p. 319) b0 = r(r + 1)a0, b1 = (r + 1)(r + 2)a1 + 3(r + 1)(r + 2)a0, bn = (n + r)(n + r + 1)an + 3(n + r)(n + r + 1)an−1 + (n + r)an−2, n ≥2 7.1.25 (p. 319) b0 = (r + 2)(r + 1)a0 b1 = (r + 3)(r + 2)a1, bn = (n + r + 2)(n + r + 1)an + 2(n + r −1)(n + r −3)an−2, n ≥2 7.1.26 (p. 319) b0 = 2(r + 1)(r + 3)a0, b1 = 2(r + 2)(r + 4)a1, bn = 2(n + r + 1)(n + r + 3)an + (n + r −3)(n + r)an−2, n ≥2 Section 7.2 Answers, pp. 328–333 7.2.1 (p. 328) y = a0 ∞ X m=0 (−1)m(2m + 1)x2m + a1 ∞ X m=0 (−1)m(m + 1)x2m+1 7.2.2 (p. 328) y = a0 ∞ X m=0 (−1)m+1 x2m 2m −1 + a1x 7.2.3 (p. 328) y = a0(1 −10x2 + 5x4) + a1 x −2x3 + 1 5x5 7.2.4 (p. 328) y = a0 ∞ X m=0 (m + 1)(2m + 1)x2m + a1 3 ∞ X m=0 (m + 1)(2m + 3)x2m+1 7.2.5 (p. 328) y = a0 ∞ X m=0 (−1)m " m−1 Y j=0 | Elementary Differential Equations with Boundary Value Problems_Page_743_Chunk3148 |
4j + 1 2j + 1 # x2m + a1 ∞ X m=0 (−1)m " m−1 Y j=0 (4j + 3) # x2m+1 2mm! | Elementary Differential Equations with Boundary Value Problems_Page_743_Chunk3149 |
734 Answers to Selected Exercises 7.2.6 (p. 328) y = a0 ∞ X m=0 (−1)m "m−1 Y j=0 (4j + 1)2 2j + 1 # x2m 8mm! + a1 ∞ X m=0 (−1)m "m−1 Y j=0 (4j + 3)2 2j + 3 # x2m+1 8mm! 7.2.7 (p. 328) y = a0 ∞ X m=0 2mm! Qm−1 j=0 (2j + 1) x2m + a1 ∞ X m=0 Qm−1 j=0 (2j + 3) 2mm! x2m+1 7.2.8 (p. 328) y = a0 1 −14x2 + 35 3 x4 + a1 x −3x3 + 3 5x5 + 1 35 x7 7.2.9 (p. 329) (a) y = a0 ∞ X m=0 (−1)m x2m Qm−1 j=0 (2j + 1) + a1 ∞ X m=0 (−1)m x2m+1 2mm! 7.2.10 (p. 329) (a) y = a0 ∞ X m=0 (−1)m "m−1 Y j=0 4j + 3 2j + 1 # x2m 2mm! + a1 ∞ X m=0 (−1)m "m−1 Y j=0 4j + 5 2j + 3 # x2m+1 2mm! 7.2.11 (p. 329) y = 2 −x −x2 + 1 3x3 + 5 12x4 −1 6x5 −17 72 x6 + 13 126x7 + · · · 7.2.12 (p. 329) y = 1 −x + 3x2 −5 2x3 + 5x4 −21 8 x5 + 3x6 −11 16x7 + · · · 7.2.13 (p. 329) y = 2 −x −2x2 + 1 3x3 + 3x4 −5 6x5 −49 5 x6 + 45 14x7 + · · · 7.2.16 (p. 330) y = a0 ∞ X m=0 (x −3)2m (2m)! + a1 ∞ X m=0 (x −3)2m+1 (2m + 1)! 7.2.17 (p. 330) y = a0 ∞ X m=0 (x −3)2m 2mm! + a1 ∞ X m=0 (x −3)2m+1 Qm−1 j=0 (2j + 3) 7.2.18 (p. 330) y = a0 ∞ X m=0 "m−1 Y j=0 (2j + 3) # (x −1)2m m! + a1 ∞ X m=0 4m(m + 1)! Qm−1 j=0 (2j + 3) (x −1)2m+1 7.2.19 (p. 330) y = a0 1 −6(x −2)2 + 4 3(x −2)4 + 8 135 (x −2)6 + a1 (x −2) −10 9 (x −2)3 7.2.20 (p. 330) y = a0 ∞ X m=0 (−1)m "m−1 Y j=0 (2j + 1) # 3m 4mm!(x + 1)2m + a1 ∞ X m=0 (−1)m 3mm! Qm−1 j=0 (2j + 3) (x + 1)2m+1 7.2.21 (p. 330) y = −1 + 2x + 3 8x2 −1 3x3 − 3 128 x4 − 1 1024x6 + · · · 7.2.22 (p. 330) y = −2 + 3(x −3) + 3(x −3)2 −2(x −3)3 −5 4 (x −3)4 + 3 5(x −3)5 + 7 24(x −3)6 −4 35(x −3)7 + · · · 7.2.23 (p. 330) y = −1 + (x −1) + 3(x −1)2 −5 2(x −1)3 −27 4 (x −1)4 + 21 4 (x −1)5 + 27 2 (x −1)6 −81 8 (x −1)7 + · · · 7.2.24 (p. 330) y = 4 −6(x −3) −2(x −3)2 + (x −3)3 + 3 2 (x −3)4 −5 4(x −3)5 −49 20(x −3)6 + 135 56 (x −3)7 + · · · 7.2.25 (p. 330) y = 3 −4(x −4) + 15(x −4)2 −4(x −4)3 + 15 4 (x −4)4 −1 5(x −4)5 7.2.26 (p. 330) y = 3 −3(x + 1) −30(x + 1)2 + 20 3 (x + 1)3 + 20(x + 1)4 −4 3(x + 1)5 −8 9(x + 1)6 7.2.27 (p. 330) (a)y = a0 ∞ X m=0 (−1)mx2m + a1 ∞ X m=0 (−1)mx2m+1 (b)y = a0 + a1x 1 + x2 7.2.33 (p. 333) y = a0 ∞ X m=0 x3m 3mm! Qm−1 j=0 (3j + 2) + a1 ∞ X m=0 x3m+1 3mm!Qm−1 j=0 (3j + 4) | Elementary Differential Equations with Boundary Value Problems_Page_744_Chunk3150 |
Answers to Selected Exercises 735 7.2.34 (p. 333) y = a0 ∞ X m=0 2 3 m "m−1 Y j=0 (3j + 2) # x3m m! + a1 ∞ X m=0 6mm! Qm−1 j=0 (3j + 4) x3m+1 7.2.35 (p. 333) y = a0 ∞ X m=0 (−1)m 3mm! Qm−1 j=0 (3j + 2) x3m + a1 ∞ X m=0 (−1)m "m−1 Y j=0 (3j + 4) # x3m+1 3mm! 7.2.36 (p. 333) y = a0(1 −4x3 + 4x6) + a1 ∞ X m=0 2m " m−1 Y j=0 3j −5 3j + 4 # x3m+1 7.2.37 (p. 333) y = a0 1 + 21 2 x3 + 42 5 x6 + 7 20x9 + a1 x + 4x4 + 10 7 x7 7.2.39 (p. 333) y = a0 ∞ X m=0 (−2)m "m−1 Y j=0 5j + 1 5j + 4 # x5m + a1 ∞ X m=0 −2 5 m "m−1 Y j=0 (5j + 2) # x5m+1 m! 7.2.40 (p. 333) y = a0 ∞ X m=0 (−1)m x4m 4mm! Qm−1 j=0 (4j + 3) + a1 ∞ X m=0 (−1)m x4m+1 4mm!Qm−1 j=0 (4j + 5) 7.2.41 (p. 333) y = a0 ∞ X m=0 (−1)m x7m Qm−1 j=0 (7j + 6) + a1 ∞ X m=0 (−1)m x7m+1 7mm! 7.2.42 (p. 333) y = a0 1 −9 7x8 + a1 x −7 9 x9 7.2.43 (p. 333) y = a0 ∞ X m=0 x6m + a1 ∞ X m=0 x6m+1 7.2.44 (p. 333) y = a0 ∞ X m=0 (−1)m x6m Qm−1 j=0 (6j + 5) + a1 ∞ X m=0 (−1)m x6m+1 6mm! Section 7.3 Answers, pp. 337–341 7.3.1 (p. 337) y = 2 −3x −2x2 + 7 2x3 −55 12 x4 + 59 8 x5 −83 6 x6 + 9547 336 x7 + · · · 7.3.2 (p. 337) y = −1 + 2x −4x3 + 4x4 + 4x5 −12x6 + 4x7 + · · · 7.3.3 (p. 337) y = 1 + x2 −2 3x3 + 11 6 x4 −9 5 x5 + 329 90 x6 −1301 315 x7 + · · · 7.3.4 (p. 337) y = x −x2 −7 2x3 + 15 2 x4 + 45 8 x5 −261 8 x6 + 207 16 x7 + · · · 7.3.5 (p. 337) y = 4 + 3x −15 4 x2 + 1 4x3 + 11 16 x4 −5 16x5 + 1 20x6 + 1 120 x7 + · · · 7.3.6 (p. 337) y = 7 + 3x −16 3 x2 + 13 3 x3 −23 9 x4 + 10 9 x5 −7 27 x6 −1 9x7 + · · · 7.3.7 (p. 337) y = 2 + 5x −7 4x2 −3 16x3 + 37 192 x4 − 7 192x5 − 1 1920 x6 + 19 11520 x7 + · · · 7.3.8 (p. 337) y = 1 −(x −1) + 4 3(x −1)3 −4 3(x −1)4 −4 5(x −1)5 + 136 45 (x −1)6 −104 63 (x −1)7 + · · · 7.3.9 (p. 337) y = 1 −(x + 1) + 4(x + 1)2 −13 3 (x + 1)3 + 77 6 (x + 1)4 −278 15 (x + 1)5 + 1942 45 (x + 1)6 −23332 315 (x + 1)7 + · · · 7.3.10 (p. 337) y = 2 −(x −1) −1 2(x −1)2 + 5 3(x −1)3 −19 12 (x −1)4 + 7 30(x −1)5 + 59 45(x −1)6 −1091 630 (x −1)7 + · · · 7.3.11 (p. 337) y = −2 + 3(x + 1) −1 2(x + 1)2 −2 3(x + 1)3 + 5 8 (x + 1)4 −11 30(x + 1)5 + 29 144(x + 1)6 −101 840 (x + 1)7 + · · · 7.3.12 (p. 337) y = 1 −2(x −1) −3(x −1)2 + 8(x −1)3 −4(x −1)4 −42 5 (x −1)5 + 19(x −1)6 −604 35 (x −1)7 + · · · 7.3.19 (p. 339) y = 2 −7x −4x2 −17 6 x3 −3 4x4 −9 40x5 + · · · | Elementary Differential Equations with Boundary Value Problems_Page_745_Chunk3151 |
736 Answers to Selected Exercises 7.3.20 (p. 339) y = 1 −2(x −1) + 1 2(x −1)2 −1 6(x −1)3 + 5 36 (x −1)4 − 73 1080(x −1)5 + · · · 7.3.21 (p. 339) y = 2 −(x + 2) −7 2(x + 2)2 + 4 3(x + 2)3 −1 24(x + 2)4 + 1 60 (x + 2)5 + · · · 7.3.22 (p. 339) y = 2 −2(x + 3) −(x + 3)2 + (x + 3)3 −11 12(x + 3)4 + 67 60 (x + 3)5 + · · · 7.3.23 (p. 339) y = −1 + 2x + 1 3x3 −5 12x4 + 2 5 x5 + · · · 7.3.24 (p. 339) y = 2 −3(x + 1) + 7 2(x + 1)2 −5(x + 1)3 + 197 24 (x + 1)4 −287 20 (x + 1)5 + · · · 7.3.25 (p. 339) y = −2 + 3(x + 2) −9 2(x + 2)2 + 11 6 (x + 2)3 + 5 24 (x + 2)4 + 7 20(x + 2)5 + · · · 7.3.26 (p. 339) y = 2 −4(x −2) −1 2(x −2)2 + 2 9(x −2)3 + 49 432 (x −2)4 + 23 1080 (x −2)5 + · · · 7.3.27 (p. 339) y = 1 + 2(x + 4) −1 6(x + 4)2 −10 27(x + 4)3 + 19 648 (x + 4)4 + 13 324 (x + 4)5 + · · · 7.3.28 (p. 339) y = −1 + 2(x + 1) −1 4(x + 1)2 + 1 2(x + 1)3 −65 96(x + 1)4 + 67 80 (x + 1)5 + · · · 7.3.31 (p. 341) (a) y = c1 1 + x + c2 1 + 2x (b) y = c1 1 −2x + c2 1 −3x (c) y = c1 1 −2x + c2x (1 −2x)2 (d) y = c1 2 + x + c2x (2 + x)2 (e) y = c1 2 + x + c2 2 + 3x 7.3.32 (p. 341) y = 1 −2x −3 2x2 + 5 3x3 + 17 24 x4 −11 20x5 + · · · 7.3.33 (p. 341) y = 1 −2x −5 2x2 + 2 3x3 −3 8x4 + 1 3 x5 + · · · 7.3.34 (p. 341) y = 6 −2x + 9x2 + 2 3x3 −23 4 x4 −3 10x5 + · · · 7.3.35 (p. 341) y = 2 −5x + 2x2 −10 3 x3 + 3 2x4 −25 12x5 + · · · 7.3.36 (p. 341) y = 3 + 6x −3x2 + x3 −2x4 −17 20 x5 + · · · 7.3.37 (p. 341) y = 3 −2x −3x2 + 3 2x3 + 3 2x4 −49 80 x5 + · · · 7.3.38 (p. 341) y = −2 + 3x + 4 3x2 −x3 −19 54x4 + 13 60 x5 + · · · 7.3.39 (p. 341) y1 = ∞ X m=0 (−1)mx2m m! = e−x2, y2 = ∞ X m=0 (−1)mx2m+1 m! = xe−x2 7.3.40 (p. 341) y = −2 + 3x + x2 −1 6x3 −3 4x4 + 31 120 x5 + · · · 7.3.41 (p. 341) y = 2 + 3x −7 2x2 −5 6x3 + 41 24 x4 + 41 120x5 + · · · 7.3.42 (p. 341) y = −3 + 5x −5x2 + 23 6 x3 −23 12 x4 + 11 30x5 + · · · 7.3.43 (p. 341) y = −2 + 3(x −1) + 3 2(x −1)2 −17 12(x −1)3 −1 12 (x −1)4 + 1 8(x −1)5 + · · · 7.3.44 (p. 341) y = 2 −3(x + 2) + 1 2(x + 2)2 −1 3(x + 2)3 + 31 24 (x + 2)4 −53 120(x + 2)5 + · · · 7.3.45 (p. 341) y = 1 −2x + 3 2x2 −11 6 x3 + 15 8 x4 −71 60x5 + · · · 7.3.46 (p. 341) y = 2 −(x + 2) −7 2(x + 2)2 −43 6 (x + 2)3 −203 24 (x + 2)4 −167 30 | Elementary Differential Equations with Boundary Value Problems_Page_746_Chunk3152 |
(x + 2)5 + · · · 7.3.47 (p. 341) y = 2 −x −x2 + 7 6x3 −x4 + 89 120x5 + · · · | Elementary Differential Equations with Boundary Value Problems_Page_746_Chunk3153 |
Answers to Selected Exercises 737 7.3.48 (p. 341) y = 1 + 3 2(x −1)2 + 1 6(x −1)3 −1 8 (x −1)5 + · · · 7.3.49 (p. 341) y = 1 −2(x −3) + 1 2(x −3)2 −1 6(x −3)3 + 1 4(x −3)4 −1 6 (x −3)5 + · · · Section 7.4 Answers, pp. 346–347 7.4.1 (p. 346) y = c1x−4 + c2x−2 7.4.2 (p. 346) y = c1x + c2x7 7.4.3 (p. 346) y = x(c1 + c2 ln x) 7.4.4 (p. 346) y = x−2(c1 + c2 ln x) 7.4.5 (p. 346) y = c1 cos(ln x) + c2 sin(ln x) 7.4.6 (p. 346) y = x2[c1 cos(3 ln x) + c2 sin(3 ln x)] 7.4.7 (p. 346) y = c1x + c2 x3 7.4.8 (p. 346) y = c1x2/3 + c2x3/4 7.4.9 (p. 346) y = x−1/2(c1 + c2 ln x) 7.4.10 (p. 346) y = c1x+c2x1/3 7.4.11 (p. 346) y = c1x2+c2x1/2 7.4.12 (p. 346) y = 1 x [c1 cos(2 ln x) + c2 sin(2 ln x] 7.4.13 (p. 346) y = x−1/3(c1 + c2 ln x) 7.4.14 (p. 346) y = x [c1 cos(3 ln x) + c2 sin(3 ln x)] 7.4.15 (p. 346) y = c1x3 + c2 x2 7.4.16 (p. 346) y = c1 x + c2x1/2 7.4.17 (p. 346) y = x2(c1 + c2 ln x) 7.4.18 (p. 346) y = 1 x2 c1 cos 1 √ 2 ln x + c2 sin 1 √ 2 ln x Section 7.5 Answers, pp. 357–364 7.5.1 (p. 357) y1 = x1/2 1 −1 5x −2 35 x2 + 31 315x3 + · · · y2 = x−1 1 + x + 1 2 x2 −1 6x3 + · · · ; 7.5.2 (p. 357) y1 = x1/3 1 −2 3x + 8 9x2 −40 81 x3 + · · · ; y2 = 1 −x + 6 5x2 −4 5 x3 + · · · 7.5.3 (p. 357) y1 = x1/3 1 −4 7x −7 45 x2 + 970 2457x3 + · · · ; y2 = x−1 1 −x2 + 2 3 x3 + · · · 7.5.4 (p. 357) y1 = x1/4 1 −1 2x −19 104 x2 + 1571 10608 x3 + · · · ; y2 = x−1 1 + 2x −11 6 x2 −1 7x3 + · · · 7.5.5 (p. 357) y1 = x1/3 1 −x + 28 31x2 −1111 1333 x3 + · · · ; y2 = x−1/4 1 −x + 7 8x2 −19 24 x3 + · · · ; 7.5.6 (p. 357) y1 = x1/5 1 −6 25 x −1217 625 x2 + 41972 46875 x3 + · · · ; y2 = x −1 4 x2 −35 18x3 + 11 12x4 + · · · 7.5.7 (p. 357) y1 = x3/2 1 −x + 11 26x2 −109 1326 x3 + · · · ; y2 = x1/4 1 + 4x −131 24 x2 + 39 14x3 + · · · 7.5.8 (p. 357) y1 = x1/3 1 −1 3x + 2 15 x2 −5 63x3 + · · · ; y2 = x−1/6 1 −1 12x2 + 1 18 x3 + · · · 7.5.9 (p. 357) y1 = 1 −1 14x2 + 1 105 x3 + · · ·; y2 = x−1/3 1 −1 18x −71 405 x2 + 719 34992 x3 + · · · 7.5.10 (p. 358) y1 = x1/5 1 + 3 17x − 7 153x2 −547 5661 x3 + · · · ; y2 = x−1/2 1 + x + 14 13x2 −556 897 x3 + · · · 7.5.14 (p. 358) y1 = x1/2 ∞ X n=0 (−2)n Qn j=1(2j + 3)xn; y2 = x−1 ∞ X n=0 (−1)n n! xn 7.5.15 (p. 358) y1 = x1/3 ∞ X n=0 (−1)n Qn j=1(3j + 1) 9nn! | Elementary Differential Equations with Boundary Value Problems_Page_747_Chunk3154 |
xn; x−1 7.5.16 (p. 358) y1 = x1/2 ∞ X n=0 (−1)n 2nn! xn; y2 = 1 x2 ∞ X n=0 (−1)n Qn j=1(2j −5)xn 7.5.17 (p. 358) y1 = x ∞ X n=0 (−1)n Qn j=1(3j + 4)xn; y2 = x−1/3 ∞ X n=0 (−1)n 3nn! xn 7.5.18 (p. 358) y1 = x ∞ X n=0 2n n! Qn j=1(2j + 1)xn; y2 = x1/2 ∞ X n=0 2n n! Qn j=1(2j −1)xn | Elementary Differential Equations with Boundary Value Problems_Page_747_Chunk3155 |
738 Answers to Selected Exercises 7.5.19 (p. 358) y1 = x1/3 ∞ X n=0 1 n! Qn j=1(3j + 2)xn; y2 = x−1/3 ∞ X n=0 1 n! Qn j=1(3j −2)xn 7.5.20 (p. 358) y1 = x 1 + 2 7x + 1 70x2 ; y2 = x−1/3 ∞ X n=0 (−1)n 3nn! n Y j=1 3j −13 3j −4 ! xn 7.5.21 (p. 358) y1 = x1/2 ∞ X n=0 (−1)n n Y j=1 2j + 1 6j + 1 ! ; xn y2 = x1/3 ∞ X n=0 (−1)n 9nn! n Y j=1 (3j + 1) ! xn 7.5.22 (p. 358) y1 = x ∞ X n=0 (−1)n(n + 2)! 2 Qn j=1(4j + 3); xn y2 = x1/4 ∞ X n=0 (−1)n 16nn! n Y j=1 (4j + 5)xn 7.5.23 (p. 358) y1 = x−1/2 ∞ X n=0 (−1)n n! Qn j=1(2j + 1)xn; y2 = x−1 ∞ X n=0 (−1)n n! Qn j=1(2j −1)xn 7.5.24 (p. 358) y1 = x1/3 ∞ X n=0 (−1)n n! 2 9 n n Y j=1 (6j + 5) ! xn; y2 = x−1 ∞ X n=0 (−1)n2n n Y j=1 2j −1 3j −4 ! xn 7.5.25 (p. 358) y1 = 4x1/3 ∞ X n=0 1 6nn!(3n + 4)xn; x−1 7.5.28 (p. 359) y1 = x1/2 1 −9 40x + 5 128x2 − 245 39936 x3 + · · · ; y2 = x1/4 1 −25 96x + 675 14336 x2 − 38025 5046272 x3 + · · · 7.5.29 (p. 359) y1 = x1/3 1 + 32 117 x − 28 1053x2 + 4480 540189 x3 + · · · ; y2 = x−3 1 + 32 7 x + 48 7 x2 7.5.30 (p. 359) y1 = x1/2 1 −5 8x + 55 96x2 −935 1536 x3 + · · · ; y2 = x−1/2 1 + 1 4x −5 32x2 −55 384 x3 + · · · . 7.5.31 (p. 359) y1 = x1/2 1 −3 4 x + 5 96 x2 + 5 4224x3 + · · · ; y2 = x−2 | Elementary Differential Equations with Boundary Value Problems_Page_748_Chunk3156 |
Answers to Selected Exercises 739 7.5.40 (p. 359) y1 = x1/2 ∞ X m=0 (−1)m 4m m Y j=1 4j −1 2j + 1 ! x2m; y2 = x−1/2 ∞ X m=0 (−1)m 8mm! m Y j=1 (4j −3) ! x2m 7.5.41 (p. 359) y1 = x1/2 ∞ X m=0 (−1)m m! m Y j=1 (2j + 1) ! x2m; y2 = 1 x2 ∞ X m=0 (−2)m m Y j=1 4j −3 4j −5 ! x2m 7.5.42 (p. 359) y1 = x1/3 ∞ X m=0 (−1)m m Y j=1 3j −4 3j + 2 ! x2m; y2 = x−1(1 + x2) 7.5.43 (p. 359) y1 = ∞ X m=0 (−1)m 2m(m + 1)! Qm j=1(2j + 3)x2m; y2 = 1 x3 ∞ X m=0 (−1)m Qm j=1(2j −1) 2mm! x2m 7.5.44 (p. 359) y1 = x1/2 ∞ X m=0 (−1)m 8mm! m Y j=1 (4j −3)2 4j + 3 ! x2m; y2 = x−1 ∞ X m=0 (−1)m 2mm! m Y j=1 (2j −3)2 4j −3 ! x2m 7.5.45 (p. 359) y1 = x ∞ X m=0 (−2)m m Y j=1 2j + 1 4j + 5 ! x2m; y2 = x−3/2 ∞ X m=0 (−1)m 4mm! m Y j=1 (4j −3) ! x2m 7.5.46 (p. 359) y1 = x1/3 ∞ X m=0 (−1)m 2m Qm j=1(3j + 1) x2m; y2 = x−1/3 ∞ X m=0 (−1)m 6mm! x2m 7.5.47 (p. 359) y1 = x1/2 1 −6 13x2 + 36 325 x4 − 216 12025 x6 + · · · ; y2 = x1/3 1 −1 2x2 + 1 8x4 −1 48x6 + · · · 7.5.48 (p. 359) y1 = x1/4 1 −13 64x2 + 273 8192 x4 − 2639 524288 x6 + · · · ; y2 = x−1 1 −1 3x2 + 2 33 x4 − 2 209x6 + · · · 7.5.49 (p. 359) y1 = x1/3 1 −3 4x2 + 9 14x4 −81 140 x6 + · · · ; y2 = x−1/3 1 −2 3x2 + 5 9x4 −40 81 x6 + · · · 7.5.50 (p. 359) y1 = x1/2 1 −3 2 x2 + 15 8 x4 −35 16x6 + · · · ; y2 = x−1/2 1 −2x2 + 8 3x4 −16 5 x6 + · · · 7.5.51 (p. 359) y1 = x1/4 1 −x2 + 3 2 x4 −5 2x6 + · · · ; y2 = x−1/2 1 −2 5x2 + 36 65x4 −408 455 x6 + · · · 7.5.53 (p. 360) (a) y1 = xν ∞ X m=0 (−1)m 4mm! Qm j=1(j + ν) x2m; y2 = x−ν ∞ X m=0 (−1)m 4mm! Qm j=1(j −ν)x2m y1 = sin x √x ; y2 = cos x √x 7.5.61 (p. 363) y1 = x1/2 1 + x; y2 = x 1 + x 7.5.62 (p. 364) y1 = x1/3 1 + 2x2 ; y2 = x1/2 1 + 2x2 7.5.63 (p. 364) y1 = x1/4 1 −3x; y2 = x2 1 −3x 7.5.64 (p. 364) y1 = x1/3 5 + x; y2 = x−1/3 5 + x 7.5.65 (p. 364) y1 = x1/4 2 −x2 ; y2 = x−1/2 2 −x2 7.5.66 (p. 364) y1 = x1/2 1 + 3x + x2 ; y2 = x3/2 1 + 3x + x2 7.5.67 (p. 364) y1 = x (1 + x)2 ; y2 = x1/3 (1 + x)2 7.5.68 (p. 364) y1 = x 3 + 2x + x2 ; y2 = x1/4 3 + 2x + x2 Section 7.6 Answers, pp. 373–378 7.6.1 (p. 373) y1 = x 1 −x + 3 4x2 −13 36x3 + · · · ; y2 = y1 ln x + x2 1 −x + 65 108 x2 + · · · 7.6.2 (p. 373) y1 = x−1 1 −2x + 9 2x2 −20 3 x3 + · · · ; y2 = y1 ln x + 1 −15 4 x + 133 18 x2 + · · | Elementary Differential Equations with Boundary Value Problems_Page_749_Chunk3157 |
· 7.6.3 (p. 373) y1 = 1 + x −x2 + 1 3x3 + · · ·; y2 = y1 ln x −x 3 −1 2 x −31 18 x2 + · · · | Elementary Differential Equations with Boundary Value Problems_Page_749_Chunk3158 |
740 Answers to Selected Exercises 7.6.4 (p. 373) y1 = x1/2 1 −2x + 5 2x2 −2x3 + · · · ; y2 = y1 ln x + x3/2 1 −9 4x + 17 6 x2 + · · · 7.6.5 (p. 373) y1 = x 1 −4x + 19 2 x2 −49 3 x3 + · · · ; y2 = y1 ln x + x2 3 −43 4 x + 208 9 x2 + · · · 7.6.6 (p. 373) y1 = x−1/3 1 −x + 5 6x2 −1 2x3 + · · · ; y2 = y1 ln x + x2/3 1 −11 12x + 25 36x2 + · · · 7.6.7 (p. 373) y1 = 1 −2x + 7 4x2 −7 9x3 + · · ·; y2 = y1 ln x + x 3 −15 4 x + 239 108x2 + · · · 7.6.8 (p. 373) y1 = x−2 1 −2x + 5 2x2 −3x3 + · · · ; y2 = y1 ln x + 3 4 −13 6 x + · · · 7.6.9 (p. 373) y1 = x−1/2 1 −x + 1 4x2 + 1 18 x3 + · · · ; y2 = y1 ln x + x1/2 3 2 −13 16 x + 1 54 x2 + · · · 7.6.10 (p. 373) y1 = x−1/4 1 −1 4x −7 32x2 + 23 384x3 + · · · ; y2 = y1 ln x + x3/4 1 4 + 5 64x −157 2304x2 + · · · 7.6.11 (p. 373) y1 = x−1/3 1 −x + 7 6x2 −23 18 x3 + · · · ; y2 = y1 ln x −x5/3 1 12 −13 108x · · · 7.6.12 (p. 373) y1 = x1/2 ∞ X n=0 (−1)n (n!)2 xn; y2 = y1 ln x −2x1/2 ∞ X n=1 (−1)n (n!)2 n X j=1 1 j ! xn; 7.6.13 (p. 374) y1 = x1/6 ∞ X n=0 2 3 n Qn j=1(3j + 1) n! xn; y2 = y1 ln x −x1/6 ∞ X n=1 2 3 n Qn j=1(3j + 1) n! n X j=1 1 j(3j + 1) ! xn 7.6.14 (p. 374) y1 = x2 ∞ X n=0 (−1)n(n + 1)2xn; y2 = y1 ln x −2x2 ∞ X n=1 (−1)nn(n + 1)xn 7.6.15 (p. 374) y1 = x3 ∞ X n=0 2n(n + 1)xn; y2 = y1 ln x −x3 ∞ X n=1 2nnxn 7.6.16 (p. 374) y1 = x1/5 ∞ X n=0 (−1)n Qn j=1(5j + 1) 125n(n!)2 xn; y2 = y1 ln x −x1/5 ∞ X n=1 (−1)n Qn j=1(5j + 1) 125n(n!)2 n X j=1 5j + 2 j(5j + 1) ! xn 7.6.17 (p. 374) y1 = x1/2 ∞ X n=0 (−1)n Qn j=1(2j −3) 4nn! xn; y2 = y1 ln x + 3x1/2 ∞ X n=1 (−1)n Qn j=1(2j −3) 4nn! n X j=1 1 j(2j −3) ! xn 7.6.18 (p. 374) y1 = x1/3 ∞ X n=0 (−1)n Qn j=1(6j −7)2 81n(n!)2 xn; y2 = y1 ln x + 14x1/3 ∞ X n=1 (−1)n Qn j=1(6j −7)2 81n(n!)2 n X j=1 1 j(6j −7)) ! xn 7.6.19 (p. 374) y1 = x2 ∞ X n=0 (−1)n Qn j=1(2j + 5) (n!)2 xn; y2 = y1 ln x −2x2 ∞ X n=1 (−1)n Qn j=1(2j + 5) (n!)2 n X j=1 (j + 5) j(2j + 5) ! xn | Elementary Differential Equations with Boundary Value Problems_Page_750_Chunk3159 |
Answers to Selected Exercises 741 7.6.20 (p. 374) y1 = 1 x ∞ X n=0 2n Qn j=1(2j −1) n! xn; y2 = y1 ln x + 1 x ∞ X n=1 2n Qn j=1(2j −1) n! n X j=1 1 j(2j −1) ! xn 7.6.21 (p. 374) y1 = 1 x ∞ X n=0 (−1)n Qn j=1(2j −5) n! xn; y2 = y1 ln x + 5 x ∞ X n=1 (−1)n Qn j=1(2j −5) n! n X j=1 1 j(2j −5) ! xn 7.6.22 (p. 374) y1 = x2 ∞ X n=0 (−1)n Qn j=1(2j + 3) 2nn! xn; y2 = y1 ln x −3x2 ∞ X n=0 (−1)n Qn j=1(2j + 3) 2nn! n X j=1 1 j(2j + 3) ! xn 7.6.23 (p. 374) y1 = x−2 1 + 3x + 3 2x2 −1 2x3 + · · · ; y2 = y1 ln x −5x−1 1 + 5 4x −1 4x2 + · · · 7.6.24 (p. 374) y1 = x3(1 + 20x + 180x2 + 1120x3 + · · ·; y2 = y1 ln x −x4 26 + 324x + 6968 3 x2 + · · · 7.6.25 (p. 374) y1 = x 1 −5x + 85 4 x2 −3145 36 x3 + · · · ; y2 = y1 ln x + x2 2 −39 4 x + 4499 108 x2 + · · · 7.6.26 (p. 374) y1 = 1 −x + 3 4x2 −7 12x3 + · · ·; y2 = y1 ln x + x 1 −3 4 x + 5 9x2 + · · · 7.6.27 (p. 374) y1 = x−3(1 + 16x + 36x2 + 16x3 + · · · ); y2 = y1 ln x −x−2 40 + 150x + 280 3 x2 + · · · 7.6.28 (p. 374) y1 = x ∞ X m=0 (−1)m 2mm! x2m; y2 = y1 ln x −x 2 ∞ X m=1 (−1)m 2mm! m X j=1 1 j ! x2m 7.6.29 (p. 374) y1 = x2 ∞ X m=0 (−1)m(m + 1)x2m; y2 = y1 ln x −x2 2 ∞ X m=1 (−1)mmx2m 7.6.30 (p. 374) y1 = x1/2 ∞ X m=0 (−1)m 4mm! x2m; y2 = y1 ln x −x1/2 2 ∞ X m=1 (−1)m 4mm! m X j=1 1 j ! x2m 7.6.31 (p. 374) y1 = x ∞ X m=0 (−1)m Qm j=1(2j −1) 2mm! x2m; y2 = y1 ln x + x 2 ∞ X m=1 (−1)m Qm j=1(2j −1) 2mm! m X j=1 1 j(2j −1) ! x2m 7.6.32 (p. 374) y1 = x1/2 ∞ X m=0 (−1)m Qm j=1(4j −1) 8mm! x2m; y2 = y1 ln x + x1/2 2 ∞ X m=1 (−1)m Qm j=1(4j −1) 8mm! m X j=1 1 j(4j −1) ! x2m 7.6.33 (p. 374) y1 = x ∞ X m=0 (−1)m Qm j=1(2j + 1) 2mm! x2m; y2 = y1 ln x −x 2 ∞ X m=1 (−1)m Qm j=1(2j + 1) 2mm! m X j=1 1 j(2j + 1) ! x2m | Elementary Differential Equations with Boundary Value Problems_Page_751_Chunk3160 |
742 Answers to Selected Exercises 7.6.34 (p. 374) y1 = x−1/4 ∞ X m=0 (−1)m Qm j=1(8j −13) (32)mm! x2m; y2 = y1 ln x + 13 2 x−1/4 ∞ X m=1 (−1)m Qm j=1(8j −13) (32)mm! m X j=1 1 j(8j −13) ! x2m 7.6.35 (p. 374) y1 = x1/3 ∞ X m=0 (−1)m Qm j=1(3j −1) 9mm! x2m; y2 = y1 ln x + x1/3 2 ∞ X m=1 (−1)m Qm j=1(3j −1) 9mm! m X j=1 1 j(3j −1) ! x2m 7.6.36 (p. 374) y1 = x1/2 ∞ X m=0 (−1)m Qm j=1(4j −3)(4j −1) 4m(m!)2 x2m; y2 = y1 ln x + x1/2 ∞ X m=1 (−1)m Qm j=1(4j −3)(4j −1) 4m(m!)2 m X j=1 8j −3 j(4j −3)(4j −1) ! x2m 7.6.37 (p. 374) y1 = x5/3 ∞ X m=0 (−1)m 3mm! x2m; y2 = y21 ln x −x5/3 2 ∞ X m=1 (−1)m 3mm! m X j=1 1 j ! x2m 7.6.38 (p. 374) y1 = 1 x ∞ X m=0 (−1)m Qm j=1(4j −7) 2mm! x2m; y2 = y1 ln x + 7 2x ∞ X m=1 (−1)m Qm j=1(4j −7) 2mm! m X j=1 1 j(4j −7) ! x2m 7.6.39 (p. 374) y1 = x−1 1 −3 2x2 + 15 8 x4 −35 16x6 + · · · ; y2 = y1 ln x + x 1 4 −13 32 x2 + 101 192x4 + · · · 7.6.40 (p. 375) y1 = x 1 −1 2x2 + 1 8x4 −1 48 x6 + · · · ; y2 = y1 ln x + x3 1 4 −3 32 x2 + 11 576x4 + · · · 7.6.41 (p. 375) y1 = x−2 1 −3 4x2 −9 64 x4 −25 256x6 + · · · ; y2 = y1 ln x + 1 2 −21 128x2 −215 1536 x4 + · · · 7.6.42 (p. 375) y1 = x−3 1 −17 8 x2 + 85 256x4 − 85 18432 x6 + · · · ; y2 = y1 ln x + x−1 25 8 −471 512x2 + 1583 110592 x4 + · · · 7.6.43 (p. 375) y1 = x−1 1 −3 4x2 + 45 64x4 −175 256 x6 + · · · ; y2 = y1 ln x −x 1 4 −33 128 x2 + 395 1536 x4 + · · · 7.6.44 (p. 375) y1 = 1 x; y2 = y1 ln x −6 + 6x −8 3x2 7.6.45 (p. 375) y1 = 1 −x; y2 = y1 ln x + 4x 7.6.46 (p. 375) y1 = (x −1)2 x ; y2 = y1 ln x + 3 −3x + 2 ∞ X n=2 1 n(n2 −1)xn 7.6.47 (p. 375) y1 = x1/2(x + 1)2; y2 = y1 ln x −x3/2 3 + 3x + 2 ∞ X n=2 (−1)n n(n2 −1)xn ! 7.6.48 (p. 375) y1 = x2(1 −x)3; y2 = y1 ln x + x3 4 −7x + 11 3 x2 −6 ∞ X n=3 1 n(n −2)(n2 −1)xn ! 7.6.49 (p. 375) y1 = x −4x3 + x5; y2 = y1 ln x + 6x3 −3x5 7.6.50 (p. 375) y1 = x1/3 1 −1 6 x2 ; y2 = y1 ln x + x7/3 1 4 −1 12 ∞ X m=1 1 6mm(m + 1)(m + 1)!x2m ! | Elementary Differential Equations with Boundary Value Problems_Page_752_Chunk3161 |
Answers to Selected Exercises 743 7.6.51 (p. 375) y1 = (1 + x2)2; y2 = y1 ln x −3 2x2 −3 2x4 + ∞ X m=3 (−1)m m(m −1)(m −2)x2m 7.6.52 (p. 375) y1 = x−1/2 1 −1 2x2 + 1 32x4 ; y2 = y1 ln x + x3/2 5 8 − 9 128 x2 + ∞ X m=2 1 4m+1(m −1)m(m + 1)(m + 1)!x2m ! . 7.6.56 (p. 377) y1 = ∞ X m=0 (−1)m 4m(m!)2 x2m; y2 = y1 ln x − ∞ X m=1 (−1)m 4m(m!)2 m X j=1 1 j ! x2m 7.6.58 (p. 378) x1/2 1 + x; x1/2 ln x 1 + x 7.6.59 (p. 378) x1/3 3 + x; x1/3 ln x 3 + x 7.6.60 (p. 378) x 2 −x2 ; x ln x 2 −x2 7.6.61 (p. 378) x1/4 1 + x2 ; x1/4 ln x 1 + x2 7.6.62 (p. 378) x 4 + 3x; x ln x 4 + 3x 7.6.63 (p. 378) x1/2 1 + 3x + x2 ; x1/2 ln x 1 + 3x + x2 7.6.64 (p. 378) x (1 −x)2 ; x ln x (1 −x)2 7.6.65 (p. 378) x1/3 1 + x + x2 ; x1/3 ln x 1 + x + x2 | Elementary Differential Equations with Boundary Value Problems_Page_753_Chunk3162 |
744 Answers to Selected Exercises Section 7.7 Answers, pp. 388–390 7.7.1 (p. 388) y1 = 2x3 ∞ X n=0 (−4)n n!(n + 2)!xn; y2 = x + 4x2 −8 y1 ln x −4 ∞ X n=1 (−4)n n!(n + 2)! n X j=1 j + 1 j(j + 2) ! xn ! 7.7.2 (p. 388) y1 = x ∞ X n=0 (−1)n n!(n + 1)!xn; y2 = 1 −y1 ln x + x ∞ X n=1 (−1)n n!(n + 1)! n X j=1 2j + 1 j(j + 1) ! xn 7.7.3 (p. 388) y1 = x1/2; y2 = x−1/2 + y1 ln x + x1/2 ∞ X n=1 (−1)n n xn 7.7.4 (p. 388) y1 = x ∞ X n=0 (−1)n n! xn = xe−x; y2 = 1 −y1 ln x + x ∞ X n=1 (−1)n n! n X j=1 1 j ! xn 7.7.5 (p. 388) y1 = x1/2 ∞ X n=0 −3 4 n Qn j=1(2j + 1) n! xn; y2 = x−1/2 −3 4 y1 ln x −x1/2 ∞ X n=1 −3 4 n Qn j=1(2j + 1) n! n X j=1 1 j(2j + 1) ! xn ! 7.7.6 (p. 388) y1 = x ∞ X n=0 (−1)n n! xn = xe−x; y2 = x−2 1 + 1 2 x + 1 2x2 −1 2 y1 ln x −x ∞ X n=1 (−1)n n! n X j=1 1 j ! xn ! 7.7.7 (p. 388) y1 = 6x3/2 ∞ X n=0 (−1)n 4nn!(n + 3)!xn; y2 = x−3/2 1 + 1 8x + 1 64x2 − 1 768 y1 ln x −6x3/2 ∞ X n=1 (−1)n 4nn!(n + 3)! n X j=1 2j + 3 j(j + 3) ! xn ! 7.7.8 (p. 388) y1 = 120 x2 ∞ X n=0 (−1)n n!(n + 5)!xn; y2 = x−7 1 + 1 4 x + 1 24 x2 + 1 144x3 + 1 576 x4 − 1 2880 y1 ln x −120 x2 ∞ X n=1 (−1)n n!(n + 5)! n X j=1 2j + 5 j(j + 5) ! xn ! 7.7.9 (p. 388) y1 = x1/2 6 ∞ X n=0 (−1)n(n + 1)(n + 2)(n + 3)xn; y2 = x−5/2 1 + 1 2x + x2 −3y1 ln x + 3 2x1/2 ∞ X n=1 (−1)n(n + 1)(n + 2)(n + 3) n X j=1 1 j(j + 3) ! xn 7.7.10 (p. 388) y1 = x4 1 −2 5x y2 = 1 + 10x + 50x2 + 200x3 −300 y1 ln x + 27 25x5 −1 30 x6 7.7.11 (p. 388) y1 = x3; y2 = x−3 1 −6 5x + 3 4x2 −1 3x3 + 1 8x4 −1 20x5 − 1 120 y1 ln x + x3 ∞ X n=1 (−1)n6! n(n + 6)!xn ! 7.7.12 (p. 388) y1 = x2 ∞ X n=0 1 n! n Y j=1 2j + 3 j + 4 ! xn; y2 = x−2 1 + x + 1 4 x2 −1 12x3 −1 16y1 ln x + x2 8 ∞ X n=1 1 n! n Y j=1 2j + 3 j + 4 ! n X j=1 (j2 + 3j + 6) j(j + 4)(2j + 3) ! xn | Elementary Differential Equations with Boundary Value Problems_Page_754_Chunk3163 |
Answers to Selected Exercises 745 7.7.13 (p. 388) y1 = x5 ∞ X n=0 (−1)n(n + 1)(n + 2)xn; y2 = 1 −x 2 + x2 6 7.7.14 (p. 388) y1 = 1 x ∞ X n=0 (−1)n n! n Y j=1 (j + 3)(2j −3) j + 6 ! xn; y2 = x−7 1 + 26 5 x + 143 20 x2 7.7.15 (p. 388) y1 = x7/2 ∞ X n=0 (−1)n 2n(n + 4)!xn; y2 = x−1/2 1 −1 2x + 1 8x2 −1 48 x3 7.7.16 (p. 388) y1 = x10/3 ∞ X n=0 (−1)n(n + 1) 9n n Y j=1 3j + 7 j + 4 ! xn; y2 = x−2/3 1 + 4 27x − 1 243x2 7.7.17 (p. 388) y1 = x3 7 X n=0 (−1)n(n + 1) n Y j=1 j −8 j + 6 ! xn; y2 = x−3 1 + 52 5 x + 234 5 x2 + 572 5 x3 + 143x4 7.7.18 (p. 388) y1 = x3 ∞ X n=0 (−1)n n! n Y j=1 (j + 3)2 j + 5 ! xn; y2 = x−2 1 + 1 4 x 7.7.19 (p. 388) y1 = x6 4 X n=0 (−1)n2n n Y j=1 j −5 j + 5 ! xn; y2 = x(1 + 18x + 144x2 + 672x3 + 2016x4) 7.7.20 (p. 388) y1 = x6 1 + 2 3x + 1 7x2 ; y2 = x 1 + 21 4 x + 21 2 x2 + 35 4 x3 7.7.21 (p. 388) y1 = x7/2 ∞ X n=0 (−1)n(n + 1)xn; y2 = x−7/2 1 −5 6x + 2 3x2 −1 2x3 + 1 3x4 −1 6x5 7.7.22 (p. 388) y1 = x10 6 ∞ X n=0 (−1)n2n(n + 1)(n + 2)(n + 3)xn; y2 = 1 −4 3x + 5 3x2 −40 21x3 + 40 21x4 −32 21 x5 + 16 21x6 7.7.23 (p. 388) y1 = x6 ∞ X m=0 (−1)m Qm j=1(2j + 5) 2mm! x2m; y2 = x2 1 + 3 2x2 −15 2 y1 ln x + 75 2 x6 ∞ X m=1 (−1)m Qm j=1(2j + 5) 2m+1m! m X j=1 1 j(2j + 5) ! x2m 7.7.24 (p. 388) y1 = x6 ∞ X m=0 (−1)m 2mm! x2m = x6e−x2/2; y2 = x2 1 + 1 2x2 −1 2y1 ln x + x6 4 ∞ X m=1 (−1)m 2mm! m X j=1 1 j ! x2m 7.7.25 (p. 388) y1 = 6x6 ∞ X m=0 (−1)m 4mm!(m + 3)!x2m; y2 = 1 + 1 8x2 + 1 64x4 − 1 384 y1 ln x −3x6 ∞ X m=1 (−1)m 4mm!(m + 3)! m X j=1 2j + 3 j(j + 3) ! x2m ! 7.7.26 (p. 388) y1 = x 2 ∞ X m=0 (−1)m(m + 2) m! x2m; y2 = x−1 −4y1 ln x + x ∞ X m=1 (−1)m(m + 2) m! m X j=1 j2 + 4j + 2 j(j + 1)(j + 2) ! x2m | Elementary Differential Equations with Boundary Value Problems_Page_755_Chunk3164 |
746 Answers to Selected Exercises 7.7.27 (p. 388) y1 = 2x3 ∞ X m=0 (−1)m 4mm!(m + 2)!x2m; y2 = x−1 1 + 1 4x2 −1 16 y1 ln x −2x3 ∞ X m=1 (−1)m 4mm!(m + 2)! m X j=1 j + 1 j(j + 2) ! x2m ! 7.7.28 (p. 388) y1 = x−1/2 ∞ X m=0 (−1)m Qm j=1(2j −1) 8mm!(m + 1)! x2m; y2 = x−5/2 + 1 4y1 ln x −x−1/2 ∞ X m=1 (−1)m Qm j=1(2j −1) 8m+1m!(m + 1)! m X j=1 2j2 −2j −1 j(j + 1)(2j −1) ! x2m 7.7.29 (p. 388) y1 = x ∞ X m=0 (−1)m 2mm! x2m = xe−x2/2; y2 = x−1 −y1 ln x + x 2 ∞ X m=1 (−1)m 2mm! m X j=1 1 j ! x2m 7.7.30 (p. 388) y1 = x2 ∞ X m=0 1 m!x2m = x2ex2; y2 = x−2(1 −x2) −2y1 ln x + x2 ∞ X m=1 1 m! m X j=1 1 j ! x2m 7.7.31 (p. 388) y1 = 6x5/2 ∞ X m=0 (−1)m 16mm!(m + 3)!x2m; y2 = x−7/2 1 + 1 32x2 + 1 1024 x4 − 1 24576 y1 ln x −3x5/2 ∞ X m=1 (−1)m 16mm!(m + 3)! m X j=1 2j + 3 j(j + 3) ! x2m ! 7.7.32 (p. 388) y1 = 2x13/3 ∞ X m=0 Qm j=1(3j + 1) 9mm!(m + 2)!x2m; y2 = x1/3 1 + 2 9x2 + 2 81 y1 ln x −x13/3 ∞ X m=0 Qm j=1(3j + 1) 9mm!(m + 2)! m X j=1 3j2 + 2j + 2 j(j + 2)(3j + 1) ! x2m ! 7.7.33 (p. 388) y1 = x2; y2 = x−2(1 + 2x2) −2 y1 ln x + x2 ∞ X m=1 1 m(m + 2)!x2m ! 7.7.34 (p. 388) y1 = x2 1 −1 2x2 ; y2 = x−2 1 + 9 2x2 −27 2 y1 ln x + 7 12 x4 −x2 ∞ X m=2 | Elementary Differential Equations with Boundary Value Problems_Page_756_Chunk3165 |
Answers to Selected Exercises 747 7.7.42 (p. 389) y1 = xν ∞ X m=0 (−1)m 4mm! Qm j=1(j + ν)x2m; y2 = x−ν ν−1 X m=0 (−1)m 4mm! Qm j=1(j −ν)x2m − 2 4νν!(ν −1)! y1 ln x −xν 2 ∞ X m=1 (−1)m 4mm!Qm j=1(j + ν) m X j=1 2j + ν j(j + ν) ! x2m ! Section 8.1 Answers, pp. 402–404 8.1.1 (p. 402) (a) 1 s2 (b) 1 (s + 1)2 (c) b s2 −b2 (d) −2s + 5 (s −1)(s −2) (e) 2 s3 8.1.2 (p. 402) (a) s2 + 2 [(s −1)2 + 1] [(s + 1)2 + 1] (b) 2 s(s2 + 4) (c) s2 + 8 s(s2 + 16) (d) s2 −2 s(s2 −4) (e) 4s (s2 −4)2 (f) 1 s2 + 4 (g) 1 √ 2 s + 1 s2 + 1 (h) 5s (s2 + 4)(s2 + 9) (i) s3 + 2s2 + 4s + 32 (s2 + 4)(s2 + 16) 8.1.4 (p. 402) (a) f(3−) = −1, f(3) = f(3+) = 1 (b) f(1−) = 3, f(1) = 4, f(1+) = 1 (c) f π 2 − = 1, f π 2 = f π 2 + = 2, f(π−) = 0, f(π) = f(π+) = −1 (d) f(1−) = 1, f(1) = 2, f(1+) = 1, f(2−) = 0, f(2) = 3, f(2+) = 6 8.1.5 (p. 402) (a) 1 −e−(s+1) s + 1 + e−(s+2) s + 2 (b) 1 s + e−4s 1 s2 + 3 s (c) 1 −e−s s2 (d) 1 −e−(s−1) (s −1)2 8.1.7 (p. 402) L(eλt cos ωt) = (s −λ)2 −ω2 ((s −λ)2 + ω2)2 L(eλt sin ωt) = 2ω(s −λ) ((s −λ)2 + ω2)2 8.1.15 (p. 403) (a) tan−1 ω s , s > 0 (b) 1 2 ln s2 s2 + ω2 , s > 0 (c) ln s −b s −a , s > max(a, b) (d) 1 2 ln s2 s2 −1, s > 1 (e) 1 4 ln s2 s2 −4, s > 2 8.1.18 (p. 404) (a) 1 s2 tanh s 2 (b) 1 s tanh s 4 (c) 1 s2 + 1 coth πs 2 (d) 1 (s2 + 1)(1 −e−πs) Section 8.2 Answers, pp. 411–413 8.2.1 (p. 411) (a) t3e7t 2 (b) 2e2t cos 3t (c) e−2t 4 sin 4t (d) 2 3 sin 3t (e) t cos t (f) e2t 2 sinh 2t (g) 2te2t 3 sin 9t (h) 2e3t 3 sinh 3t (i) e2tt cos t 8.2.2 (p. 411) (a) t2e7t + 17 6 t3e7t (b) e2t 1 6t3 + 1 6t4 + 1 40 t5 (c) e−3t cos 3t + 2 3 sin 3t (d) 2 cos 3t + 1 3 sin 3t (e) (1 −t)e−t (f) cosh 3t + 1 3 sinh 3t (g) 1 −t −t2 −1 6t3 e−t (h) et 2 cos 2t + 5 2 sin 2t (i) 1 −cos t (j) 3 cosh t + 4 sinh t (k) 3et + 4 cos 3t + 1 3 sin 3t (l) 3te−2t −2 cos 2t −3 sin 2t 8.2.3 (p. 412) (a) 1 4e2t −1 4 e−2t −e−t (b) 1 5e−4t −41 5 et + 5e3t (c) −1 2e2t −13 10e−2t −1 5e3t (d) −2 5e−4t −3 5 et (e) 3 20e2t −37 12 e−2t + 1 3et + 8 5e−3t (f) 39 10 et + 3 14 e3t + 23 105e−4t −7 3e2t 8.2.4 (p. 412) (a) 4 5e−2t −1 2e−t −3 10 cos t + 11 10 sin t (b) 2 5 sin t + 6 5 cos t + 7 5e−t sin t −6 5e−t cos t (c) 8 13 e2t −8 13e−t cos 2t + 15 26e−t sin 2t (d) 1 2tet + 3 8 et + e−2t −11 8 e−3t (e) 2 3tet + 1 9et + te−2t −1 9e−2t (f) −et + 5 2 tet + cos t −3 2 sin t 8.2.5 (p. 412) (a) 3 5 cos 2t + 1 5 sin 2t −3 5 cos 3t −2 15 sin | Elementary Differential Equations with Boundary Value Problems_Page_757_Chunk3166 |
3t (b) −4 15 cos t + 1 15 sin t + 4 15 cos 4t −1 60 sin 4t | Elementary Differential Equations with Boundary Value Problems_Page_757_Chunk3167 |
748 Answers to Selected Exercises (c) 5 3 cos t + sin t −5 3 cos 2t −1 2 sin 2t (d) −1 3 cos t 2 + 2 3 sin t 2 + 1 3 cos t −1 3 sin t (e) 1 15 cos t 4 −8 15 sin t 4 −1 15 cos 4t + 1 30 sin 4t (f) 2 5 cos t 3 −3 5 sin t 3 −2 5 cos t 2 + 2 5 sin t 2 8.2.6 (p. 412) (a) et(cos 2t + sin 2t) −e−t cos 3t + 4 3 sin 3t (b) e3t −cos 2t + 3 2 sin 2t + e−t cos 2t + 1 2 sin 2t (c) e−2t 1 8 cos t + 1 4 sin t −e2t 1 8 cos 3t −1 12 sin 3t (d) e2t cos t + 1 2 sin t −e3t cos 2t −1 4 sin 2t (e) et 1 5 cos t + 2 5 sin t −e−t 1 5 cos 2t + 2 5 sin 2t (f) et/2 −cos t + 9 8 sin t + e−t/2 cos t −1 8 sin t 8.2.7 (p. 412) (a) 1−cos t (b) et 16 (1 −cos 4t) (c) 4 9e2t + 5 9e−t sin 3t −4 9e−t cos 3t (d) 3et/2 −7 2et sin 2t −3et cos 2t (e) 1 4e3t −1 4 e−t cos 2t (f) 1 9e2t −1 9e−t cos 3t + 5 9e−t sin 3t 8.2.8 (p. 412) (a) −3 10 sin t + 2 5 cos t −3 4et + 7 20e3t (b) −3 5e−t sin t + 1 5 e−t cos t −1 2e−t + 3 10 et (c) −1 10et sin t −7 10et cos t + 1 5e−t + 1 2e2t (d) −1 2et + 7 10e−t −1 5 cos 2t + 3 5 sin 2t (e) 3 10 + 1 10 e2t + 1 10et sin 2t −2 5et cos 2t (f) −4 9e2t cos 3t + 1 3e2t sin 3t −5 9e2t + et 8.2.9 (p. 413) 1 a e b a tf t a Section 8.3 Answers, pp. 418–419 8.3.1 (p. 418) y = 1 6et −9 2e−t + 16 3 e−2t 8.3.2 (p. 418) y = −1 3 + 8 15 e3t + 4 5 e−2t 8.3.3 (p. 418) y = −23 15e−2t + 1 3et + 1 5e3t 8.3.4 (p. 418) y = −1 4e2t + 17 20 e−2t + 2 5e3t 8.3.5 (p. 418) y = 11 15e−2t + 1 6et + 1 10e3t 8.3.6 (p. 418) y = et + 2e−2t −2e−t 8.3.7 (p. 418) y = 5 3 sin t −1 3 sin 2t 8.3.8 (p. 418) y = 4et −4e2t + e3t 8.3.9 (p. 418) y = −7 2e2t + 13 3 et + 1 6e4t 8.3.10 (p. 418) y = 5 2 et −4e2t + 1 2 e3t 8.3.11 (p. 418) y = 1 3et −2e−t + 5 3e−2t 8.3.12 (p. 418) y = 2 −e−2t + et 8.3.13 (p. 418) y = 1 −cos 2t + 1 2 sin 2t 8.3.14 (p. 418) y = −1 3 + 8 15 e3t + 4 5e−2t 8.3.15 (p. 418) y = 1 6et −2 3e−2t + 1 2e−t 8.3.16 (p. 418) y = −1 + et + e−t 8.3.17 (p. 418) y = cos 2t −sin 2t + sin t 8.3.18 (p. 418) y = 7 3 −7 2e−t + 1 6e3t 8.3.19 (p. 418) y = 1 + cos t 8.3.20 (p. 418) y = t + sin t 8.3.21 (p. 418) y = t −6 sin t + cos t + sin 2t 8.3.22 (p. 418) y = e−t + 4e−2t −4e−3t 8.3.23 (p. 418) y = −3 cos t −2 sin t + e−t(2 + 5t) 8.3.24 (p. 419) y = −sin t −2 cos t + 3e3t + e−t 8.3.25 (p. 419) y = (3t + 4) sin t −(2t + 6) cos t 8.3.26 | Elementary Differential Equations with Boundary Value Problems_Page_758_Chunk3168 |
(p. 419) y = −(2t + 2) cos 2t + sin 2t + 3 cos t 8.3.27 (p. 419) y = et(cos t −3 sin t) + e3t 8.3.28 (p. 419) y = −1 + t + e−t(3 cos t −5 sin t) 8.3.29 (p. 419) y = 4 cos t −3 sin t −et(3 cos t −8 sin t) 8.3.30 (p. 419) y = e−t −2et + e−2t(cos 3t −11/3 sin 3t) 8.3.31 (p. 419) y = e−t(sin t −cos t) + e−2t(cos t + 4 sin t) 8.3.32 (p. 419) y = 1 5e2t −4 3et + 32 15e−t/2 8.3.33 (p. 419) y = 1 7e2t −2 5et/2 + 9 35e−t/3 8.3.34 (p. 419) y = e−t/2(5 cos(t/2) −sin(t/2)) + 2t −4 8.3.35 (p. 419) y = 1 17 | Elementary Differential Equations with Boundary Value Problems_Page_758_Chunk3169 |
Answers to Selected Exercises 749 8.3.36 (p. 419) y = e−t/2 10 (5t + 26) −1 5(3 cos t + sin t) 8.3.37 (p. 419) y = 1 100 | Elementary Differential Equations with Boundary Value Problems_Page_759_Chunk3170 |
750 Answers to Selected Exercises 8.4.22 (p. 428) 2 + t + u(t −1)(4 −t) + u(t −3)(t −2) = 2 + t, 0 ≤t < 1, 6, 1 ≤t < 3, t + 4, t ≥3. 8.4.23 (p. 429) 5 −t + u(t −3)(7t −15) + 3 2u(t −6)(t −6)2 = 5 −t, 0 ≤t < 3, 6t −10, 3 ≤t < 6, 44 −12t + 3 2t2, t ≥6. 8.4.24 (p. 429) u(t −π)e−2(t−π)(2 cos t −5 sin t) = ( 0, 0 ≤t < π, e−2(t−π)(2 cos t −5 sin t), t ≥π. 8.4.25 (p. 429) 1 −cos t + u(t −π/2)(3 sin t + cos t) = 1 −cos t, 0 ≤t < π 2 , 1 + 3 sin t, t ≥π 2 . 8.4.26 (p. 429) u(t −2) | Elementary Differential Equations with Boundary Value Problems_Page_760_Chunk3171 |
Answers to Selected Exercises 751 8.5.10 (p. 437) y = t −sin t −2u(t −π)(t + sin t + π cos t) 8.5.11 (p. 437) y = u(t −2) t −1 2 + e2(t−2) 2 −2et−2 8.5.12 (p. 437) y = t + sin t + cos t −u(t −2π)(3t −3 sin t −6π cos t) 8.5.13 (p. 437) y = 1 2 + 1 2e−2t −e−t + u(t −2) | Elementary Differential Equations with Boundary Value Problems_Page_761_Chunk3172 |
752 Answers to Selected Exercises Section 8.6 Answers, pp. 448–452 8.6.1 (p. 448) (a) 1 2 Z t 0 τ sin 2(t −τ) dτ (b) Z t 0 e−2τ cos 3(t −τ) dτ (c) 1 2 Z t 0 sin 2τ cos 3(t −τ) dτ or 1 3 Z t 0 sin 3τ cos 2(t −τ) dτ (d) Z t 0 cos τ sin(t −τ) dτ (e) Z t 0 eaτdτ (f) e−t Z t 0 sin(t −τ) dτ (g) e−2t Z t 0 τeτ sin(t −τ) dτ (h) e−2t 2 Z t 0 τ 2(t −τ)e3τ dτ (i) Z t 0 (t −τ)eτ cos τ dτ (j) Z t 0 e−3τ cos τ cos 2(t −τ) dτ (k) 1 4!5! Z t 0 τ 4(t −τ)5e3τ dτ (l) 1 4 Z t 0 τ 2eτ sin 2(t −τ) dτ (m) 1 2 Z t 0 τ(t −τ)2e2(t−τ) dτ (n) 1 5!6! Z t 0 (t −τ)5e2(t−τ)τ 6 dτ 8.6.2 (p. 449) (a) as (s2 + a2)(s2 + b2) (b) a (s −1)(s2 + a2) (c) as (s2 −a2)2 (d) 2ωs(s2 −ω2) (s2 + ω2)4 (e) (s −1)ω ((s −1)2 + ω2)2 (f) 2 (s −2)3(s −1)2 (g) s + 1 (s + 2)2 [(s + 1)2 + ω2] (h) 1 (s −3) ((s −1)2 −1) (i) 2 (s −2)2(s2 + 4) (j) 6 s4(s −1) (k) 3 · 6! s7 [(s + 1)2 + 9] (l) 12 s7 (m) 2 · 7! s8 [(s + 1)2 + 4] (n) 48 s5(s2 + 4) 8.6.3 (p. 449) (a) y = 2 √ 5 Z t 0 f(t −τ)e−3τ/2 sinh √ 5τ 2 dτ (b) y = 1 2 Z t 0 f(t −τ) sin 2τ dτ (c) y = Z t 0 τe−τf(t −τ) dτ (d) y(t) = −1 k sin kt + cos kt + 1 k Z t 0 f(t −τ) sin kτ dτ (e) y = −2te−3t + Z t 0 τe−3τf(t −τ) dτ (f) y = 3 2 sinh 2t + 1 2 Z t 0 f(t −τ) sinh 2τ dτ (g) y = e3t + Z t 0 (e3τ −e2τ)f(t −τ) dτ (h) y = k1 ω sin ωt + k0 cos ωt + 1 ω Z t 0 f(t −τ) sin ωτ dτ 8.6.4 (p. 449) (a) y = sin t(b) y = te−t (c) y = 1 + 2tet (d) y = t + t2 2 (e) y = 4 + 5 2t2 + 1 24t4 (f) y = 1 −t 8.6.5 (p. 450) (a) 7!8! 16! t16 (b) 13!7! 21! t21 (c) 6!7! 14! t14 (d) 1 2 (e−t + sin t −cos t) (e) 1 3(cos t −cos 2t) Section 8.7 Answers, pp. 460–461 8.7.1 (p. 460) y = 1 2e2t −4e−t + 11 2 e−2t + 2u(t −1)(e−(t−1) −e−2(t−1)) 8.7.2 (p. 460) y = 2e−2t + 5e−t + 5 3u(t −1)(e(t−1) −e−2(t−1)) 8.7.3 (p. 460) y = 1 6e2t −2 3e−t −1 2e−2t + 5 2u(t −1) sinh 2(t −1) 8.7.4 (p. 460) y = 1 8(8 cos t −5 sin t −sin 3t) −2u(t −π/2) cos t 8.7.5 (p. 460) y = 1 −cos 2t + 1 2 sin 2t + 1 2u(t −3π) sin 2t 8.7.6 (p. 460) y = 4et + 3e−t −8 + 2u(t −2) sinh(t −2) 8.7.7 (p. 460) y = 1 2et −7 2e−t + 2 + 3u(t −6)(1 −e−(t−6)) | Elementary Differential Equations with Boundary Value Problems_Page_762_Chunk3173 |
Answers to Selected Exercises 753 8.7.8 (p. 460) y = e2t + 7 cos 2t −sin 2t −1 2u(t −π/2) sin 2t 8.7.9 (p. 460) y = 1 2(1 + e−2t) + u(t −1)(e−(t−1) −e−2(t−1)) 8.7.10 (p. 460) y = 1 4et + 1 4e−t(2t −5) + 2u(t −2)(t −2)e−(t−2) 8.7.11 (p. 460) y = 1 6(2 sin t + 5 sin 2t) −1 2u(t −π/2) sin 2t 8.7.12 (p. 460) y = e−t(sin t −cos t) −e−(t−π) sin t −3u(t −2π)e−(t−2π) sin t 8.7.13 (p. 460) y = e−2t cos 3t + 4 3 sin 3t −1 3u(t −π/6)e−2(t−π/6) cos 3t −2 3u(t −π/3)e−2(t−π/3) sin 3t 8.7.14 (p. 460) y = 7 10e2t −6 5e−t/2 −1 2 + 1 5 u(t −2)(e2(t−2) −e−(t−2)/2) 8.7.15 (p. 460) y = 1 17(12 cos t + 20 sin t) + 1 34et/2(10 cos t −11 sin t) −u(t −π/2)e(2t−π)/4 cos t +u(t −π)e(t−π)/2 sin t 8.7.16 (p. 460) y = 1 3(cos t −cos 2t −3 sin t) −2u(t −π/2) cos t + 3u(t −π) sin t 8.7.17 (p. 460) y = et −e−t(1 + 2t) −5u(t −1) sinh(t −1) + 3u(t −2) sinh(t −2) 8.7.18 (p. 460) y = 1 4(et −e−t(1 + 6t)) −u(t −1)e−(t−1) + 2u(t −2)e−(t−2)) 8.7.19 (p. 460) y = 5 3 sin t −1 3 sin 2t + 1 3u(t −π)(sin 2t + 2 sin t) + u(t −2π) sin t 8.7.20 (p. 460) y = 3 4 cos 2t −1 2 sin 2t + 1 4 + 1 4u(t −π/2)(1 + cos 2t) + 1 2u(t −π) sin 2t + 3 2 u(t −3π/2) sin 2t 8.7.21 (p. 460) y = cos t −sin t 8.7.22 (p. 460) y = 1 4(8e3t −12e−2t) 8.7.23 (p. 460) y = 5(e−2t −e−t) 8.7.24 (p. 460) y = e−2t(1 + 6t) 8.7.25 (p. 460) y = 1 4e−t/2(4 −19t) 8.7.29 (p. 461) y = (−1)kmω1Re−cτ/2mδ(t −τ) if ω1τ −φ = (2k + 1)π/2(k = integer) 8.7.30 (p. 461) (a) y = (em+1 −1)(et−m −e−t) 2(e −1) , m ≤t < m + 1, (m = 0, 1, . . . ) (b) y = (m + 1) sin t, 2mπ ≤t < 2(m + 1)π, (m = 0, 1, . . . ) (c) y = e2(t−m) e2m+2 −1 e2 −1 −e(t−m) em+1 −1 e −1 , m ≤t < m + 1 (m = 0, 1, . . . ) (d) y = 0, 2mπ ≤t < (2m + 1)π, −sin t, (2m + 1)π ≤t < (2m + 2)π, (m = 0, 1,. . . ) Section 9.1 Answers, pp. 470–474 9.1.2 (p. 471) y = 2x2 −3x3 + 1 x 9.1.3 (p. 471) y = 2ex+3e−x−e2x+e−3x 9.1.4 (p. 471) yi = (x −x0)i−1 (i −1)! , 1 ≤i ≤n 9.1.5 (p. 471) (b) y1 = −1 2x3 + x2 + 1 2x, y2 = 1 3x2 −1 3x, y3 = 1 4 x3 −1 3x2 + 1 12x (c) y = k0y1 + k1y2 + k2y3 9.1.7 (p. 471) 2e−x2 9.1.8 (p. 472) √ 2K cos x 9.1.9 (p. 472) (a) W(x) = 2e3x (d) y = ex(c1+c2x+c3x2) 9.1.10 (p. 472) (a) 2 (b) −e3x(c) 4 (d) 4/x2 (e) 1 (f) 2x (g) 2/x2(h) ex(x2 −2x + 2) (i) −240/x5 (j) 6e2x(2x −1)(l) −128x 9.1.24 (p. 474) (a) y′′′ = 0 (b) xy′′′ −y′′ −xy′ + y = 0 (c) (2x −3)y′′′ −2y′′ −(2x −5)y′ = 0 (d) (x2 −2x + 2)y′′′ −x2y′′ + 2xy′ −2y = 0 (e) x3y′′′ + x2y′′ −2xy′ + 2y = 0 (f) (3x −1)y′′′ −(12x −1)y′′ + 9(x + 1)y′ −9y = 0 (g) x4y(4) + 5x3y′′′ −3x2y′′ −6xy′ + 6y = 0 | Elementary Differential Equations with Boundary Value Problems_Page_763_Chunk3174 |
754 Answers to Selected Exercises (h) x4y(4) + 3x2y′′′ −x2y′′ + 2xy′ −2y = 0 (i) (2x −1)y(4) −4xy′′′ + (5 −2x)y′′ + 4xy′ −4y = 0 (j) xy(4) −y′′′ −4xy′′ + 4y′ = 0 Section 9.2 Answers, pp. 482–487 9.2.1 (p. 482) y = ex(c1 + c2x + c3x2) 9.2.2 (p. 482) y = c1ex + c2e−x + c3 cos 3x + c4 sin 3x 9.2.3 (p. 482) y = c1ex + c2 cos 4x + c3 sin 4x 9.2.4 (p. 482) y = c1ex + c2e−x + c3e−3x/2 9.2.5 (p. 482) y = c1e−x + e−2x(c1 cos x + c2 sin x) 9.2.6 (p. 482) y = c1ex + ex/2(c2 + c3x) 9.2.7 (p. 482) y = e−x/3(c1 + c2x + c3x2) 9.2.8 (p. 482) y = c1 + c2x + c3 cos x + c4 sin x 9.2.9 (p. 482) y = c1e2x + c2e−2x + c3 cos 2x + c4 sin 2x 9.2.10 (p. 482) y = (c1 + c2x) cos √ 6x + (c3 + c4x) sin √ 6x 9.2.11 (p. 482) y = e3x/2(c1 + c2x) + e−3x/2(c3 + c4x) 9.2.12 (p. 482) y = c1e−x/2 + c2e−x/3 + c3 cos x + c4 sin x 9.2.13 (p. 482) y = c1ex +c2e−2x +c3e−x/2 +c4e−3x/2 9.2.14 (p. 482) y = ex(c1+c2x+c3 cos x+c4 sin x) 9.2.15 (p. 483) y = cos 2x −2 sin 2x + e2x 9.2.16 (p. 483) y = 2ex + 3e−x −5e−3x 9.2.17 (p. 483) y = 2ex + 3xex −4e−x 9.2.18 (p. 483) y = 2e−x cos x −3e−x sin x + 4e2x 9.2.19 (p. 483) y = 9 5e−5x/3 + ex(1 + 2x) 9.2.20 (p. 483) y = e2x(1 −3x + 2x2) 9.2.21 (p. 483) y = e3x(2 −x) + 4e−x/2 9.2.22 (p. 483) y = ex/2(1 −2x) + 3e−x/2 9.2.23 (p. 483) y = 1 8(5e2x + e−2x + 10 cos 2x + 4 sin 2x) 9.2.24 (p. 483) y = −4ex + e2x −e4x + 2e−x 9.2.25 (p. 483) y = 2ex −e−x 9.2.26 (p. 483) y = e2x + e−2x + e−x(3 cos x + sin x) 9.2.27 (p. 483) y = 2e−x/2 + cos 2x −sin 2x 9.2.28 (p. 483) (a) {ex, xex, e2x} : 1 (b) {cos 2x, sin 2x, e3x} : 26 (c) {e−x cos x, e−x sin x, ex} : 5 (d) {1, x, x2, ex} 2ex (e) {ex, e−x, cos x, sin x}8 (f) {cos x, sin x, ex cos x, ex sin x} : 5 9.2.29 (p. 483) {e−3x cos 2x, e−3x sin 2x, e2x, xe2x, 1, x, x2} 9.2.30 (p. 483) {ex, xex, ex/2, xex/2, x2ex/2, cos x, sin x} 9.2.31 (p. 483) {cos 3x, x cos 3x, x2 cos 3x, sin 3x, x sin 3x, x2 sin 3x, 1, x} 9.2.32 (p. 483) {e2x, xe2x, x2e2x, e−x, xe−x, 1} 9.2.33 (p. 483) {cos x, sin x, cos 3x, x cos 3x, sin 3x, x sin 3x, e2x} 9.2.34 (p. 483) {e2x, xe2x, e−2x, xe−2x, cos 2x, x cos 2x, sin 2x, x sin 2x} 9.2.35 (p. 483) {e−x/2 cos 2x, xe−x/2 cos 2x, x2e−x/2 cos 2x, e−x/2 sin 2x, xe−x/2 sin 2x, x2e−x/2 sin 2x} 9.2.36 (p. 483) {1, x, x2, e2x, xe2x, cos 2x, x cos 2x, sin 2x, x sin 2x} 9.2.37 (p. 483) {cos(x/2), x cos(x/2), sin(x/2), x sin(x/2), cos 2x/3 x cos(2x/3), x2 cos(2x/3), sin(2x/3), x sin(2x/3), x2 sin(2x/3)} 9.2.38 (p. 483) {e−x, e3x, ex cos 2x, ex sin 2x} 9.2.39 (p. 484) (b) e(a1+a2+···+an)x Y 1≤i<j≤n (aj −ai) 9.2.43 (p. 486) (a) ex, e−x/2 cos √ 3 2 x , e−x/2 sin √ 3 2 x (b) e−x, ex/2 cos √ 3 2 x , ex/2 sin √ 3 2 x (c) {e2x cos 2x, e2x sin 2x, e−2x cos 2x, e−2x sin 2x} (d) ex, e−x, ex/2 cos √ 3 2 x , ex/2 sin √ 3 2 x , e−x/2 cos √ 3 2 x , e−x/2 sin √ 3 2 x (e) {cos 2x, sin 2x, e− √ 3x cos x, | Elementary Differential Equations with Boundary Value Problems_Page_764_Chunk3175 |
e− √ 3x sin x, e √ 3x cos x, e √ 3x sin x} (f) 1, e2x, e3x/2 cos √ 3 2 x , e3x/2 sin √ 3 2 x , ex/2 cos √ 3 2 x , ex/2 sin √ 3 2 x | Elementary Differential Equations with Boundary Value Problems_Page_764_Chunk3176 |
Answers to Selected Exercises 755 (g) e−x, ex/2 cos √ 3 2 x , ex/2 sin √ 3 2 x , e−x/2 cos √ 3 2 x , e−x/2 sin √ 3 2 x 9.2.45 (p. 487) y = c1xr1 + c2xr2 + c3xr3 (r1, r2, r3 distinct); y = c1xr1 + (c2 + c3 ln x)xr2 (r1, r2 distinct); y = [c1 + c2 ln x + c3(ln x)2]xr1; y = c1xr1 + xλ[c2 cos(ω ln x) + c3 sin(ω ln x)] Section 9.3 Answers, pp. 494–496 9.3.1 (p. 494) yp = e−x(2+x−x2) 9.3.2 (p. 494) yp = −e−3x 4 (3−x+x2) 9.3.3 (p. 494) yp = ex(1+x−x2) 9.3.4 (p. 494) yp = e−2x(1 −5x + x2). 9.3.5 (p. 494) yp = −xex 2 (1 −x + x2 −x3) 9.3.6 (p. 494) yp = x2ex(1 + x) 9.3.7 (p. 494) yp = xe−2x 2 (2 + x) 9.3.8 (p. 494) yp = x2ex 2 (2 + x) 9.3.9 (p. 494) yp = x2e2x 2 (1+2x) 9.3.10 (p. 494) yp = x2e3x(2+x−x2) 9.3.11 (p. 494) yp = x2e4x(2+x) 9.3.12 (p. 494) yp = x3ex/2 48 (1+x) 9.3.13 (p. 494) yp = e−x(1−2x+x2) 9.3.14 (p. 494) yp = e2x(1−x) 9.3.15 (p. 494) yp = e−2x(1+x+x2 −x3) 9.3.16 (p. 494) yp = ex 3 (1−x) 9.3.17 (p. 494) yp = ex(1+x)2 9.3.18 (p. 494) yp = xex(1 + x3) 9.3.19 (p. 494) yp = xex(2 + x) 9.3.20 (p. 494) yp = xe2x 6 (1 −x2) 9.3.21 (p. 494) yp = 4xe−x/2(1 + x) 9.3.22 (p. 494) yp = xex 6 (1 + x2) 9.3.23 (p. 494) yp = x2e2x 6 (1 + x + x2) 9.3.24 (p. 494) yp = x2e2x 6 (3 + x + x2) 9.3.25 (p. 494) yp = x3ex 48 (2 + x) 9.3.26 (p. 494) yp = x3ex 6 (1 + x) 9.3.27 (p. 495) yp = −x3e−x 6 (1 −x + x2) 9.3.28 (p. 495) yp = x3e2x 12 (2 + x −x2) 9.3.29 (p. 495) yp = e−x [(1 + x) cos x + (2 −x) sin x] 9.3.30 (p. 495) yp = e−x [(1 −x) cos 2x + (1 + x) sin 2x] 9.3.31 (p. 495) yp = e2x[(1 + x −x2) cos x + (1 + 2x) sin x] 9.3.32 (p. 495) yp = ex 2 [(1 + x) cos 2x + (1 −x + x2) sin 2x] 9.3.33 (p. 495) yp = x 13 (8 cos 2x + 14 sin 2x) 9.3.34 (p. 495) yp = xex[(1 + x) cos x + (3 + x) sin x] 9.3.35 (p. 495) yp = xe2x 2 [(3 −x) cos 2x + sin 2x] 9.3.36 (p. 495) yp = −xe3x 12 (x cos 3x + sin 3x) 9.3.37 (p. 495) yp = −ex 10(cos x + 7 sin x) 9.3.38 (p. 495) yp = ex 12(cos 2x −sin 2x) 9.3.39 (p. 495) yp = xe2x cos 2x 9.3.40 (p. 495) yp = −e−x 2 [(1 + x) cos x + (2 −x) sin x] 9.3.41 (p. 495) yp = xe−x 10 (cos x + 2 sin x) 9.3.42 (p. 495) yp = xex 40 (3 cos 2x −sin 2x) 9.3.43 (p. 495) yp = xe−2x 8 [(1 −x) cos 3x + (1 + x) sin 3x] 9.3.44 (p. 495) yp = −xex 4 (1 + x) sin 2x 9.3.45 (p. 495) yp = x2e−x 4 (cos x −2 sin x) 9.3.46 (p. 495) yp = −x2e2x 32 (cos 2x −sin 2x) 9.3.47 (p. 495) yp = x2e2x 8 (1 + x) sin x 9.3.48 (p. 495) yp = 2x2ex + xe2x −cos x 9.3.49 (p. 495) yp = e2x + xex + 2x cos x 9.3.50 (p. 495) yp = 2x + x2 + 2xex −3xe−x + 4e3x 9.3.51 (p. 495) yp = xex(cos 2x−2 sin 2x)+2xe2x+1 9.3.52 (p. 495) yp = x2e−2x(1+2x)−cos 2x+sin 2x | Elementary Differential Equations with Boundary Value Problems_Page_765_Chunk3177 |
756 Answers to Selected Exercises 9.3.53 (p. 495) yp = 2x2(1 + x)e−x + x cos x −2 sin x 9.3.54 (p. 495) yp = 2xex + xe−x + cos x 9.3.55 (p. 495) yp = xex 6 (cos x + sin 2x) 9.3.56 (p. 495) yp = x2 54 [(2 + 2x)ex + 3e−2x] 9.3.57 (p. 495) yp = x 8 sinh x sin x 9.3.58 (p. 495) yp = x3(1 + x)e−x + xe−2x 9.3.59 (p. 495) yp = xex(2x2 + cos x + sin x) 9.3.60 (p. 495) y = e2x(1 + x) + c1e−x + ex(c2 + c3x) 9.3.61 (p. 495) y = e3x 1 −x −x2 2 + c1ex + e−x(c2 cos x + c3 sin x) 9.3.62 (p. 496) y = xe2x(1 + x)2 + c1ex + c2e2x + c3e3x 9.3.63 (p. 496) y = x2e−x(1 −x)2 + c1 + e−x(c2 + c3x) 9.3.64 (p. 496) y = x3ex 24 (4 + x) + ex(c1 + c2x + c3x2) 9.3.65 (p. 496) y = x2e−x 16 (1 + 2x −x2) + ex(c1 + c2x) + e−x(c3 + c4x) 9.3.66 (p. 496) y = e−2xh 1 + x 2 cos x + 3 2 −2x sin x i + c1ex + c2e−x + c3e−2x 9.3.67 (p. 496) y = −xex sin 2x + c1 + c2ex + ex(c3 cos x + c4 sin x) 9.3.68 (p. 496) y = −x2ex 16 (1 + x) cos 2x + ex [(c1 + c2x) cos 2x + (c3 + c4x) sin 2x] 9.3.69 (p. 496) y = (x2 + 2)ex −e−2x + e3x 9.3.70 (p. 496) y = e−x(1 + x + x2) + (1 −x)ex 9.3.71 (p. 496) y = x2 12 + 16 xe−x/2 −ex 9.3.72 (p. 496) y = (2 −x)(x2 + 1)e−x + cos x −sin x 9.3.73 (p. 496) y = (2 −x) cos x −(1 −7x) sin x + e−2x 9.3.74 (p. 496) 2 + ex [(1 + x) cos x −sin x −1] Section 9.4 Answers, pp. 502–505 9.4.1 (p. 502) yp = 2x3 9.4.2 (p. 503) yp = 8 105x7/2e−x2 9.4.3 (p. 503) yp = x ln |x| 9.4.4 (p. 503) yp = −2(x2 + 2) x 9.4.5 (p. 503) yp = −xe−3x 64 9.4.6 (p. 503) yp = −2x2 3 9.4.7 (p. 503) yp = −e−x(x + 1) x 9.4.8 (p. 503) yp = 2x2 ln |x| 9.4.9 (p. 503) yp = x2 + 1 9.4.10 (p. 503) yp = 2x2 + 6 3 9.4.11 (p. 503) yp = x2 ln |x| 3 9.4.12 (p. 503) yp = −x2 −2 9.4.13 (p. 503) 1 4x3 ln |x| −25 48x3 9.4.14 (p. 503) yp = x5/2 4 9.4.15 (p. 503) yp = x(12 −x2) 6 9.4.16 (p. 503) yp = x4 ln |x| 6 9.4.17 (p. 503) yp = x3ex 2 9.4.18 (p. 503) yp = x2 ln |x| 9.4.19 (p. 503) yp = xex 2 9.4.20 (p. 503) yp = 3xex 2 9.4.21 (p. 503) yp = −x3 9.4.22 (p. 503) y = −x(ln x)2 + 3x + x3 −2x ln x 9.4.23 (p. 503) y = x3 2 (ln |x|)2 + x2 −x3 + 2x3 ln |x| 9.4.24 (p. 503) y = −1 2(3x + 1)xex −3ex −e2x + 4xe−x 9.4.25 (p. 503) y = 3 2x4(ln x)2 + 3x −x4 + 2x4 ln x 9.4.26 (p. 503) y = −x4 + 12 6 + 3x −x2 + 2ex 9.4.27 (p. 503) y = x2 3 −x 2 ln |x| + 4x −2x2 9.4.28 (p. 504) y = −xex(1 + 3x) 2 + x + 1 2 −ex 4 + e3x 2 9.4.29 (p. 504) y = −8x + 2x2 −2x3 + 2ex −e−x 9.4.30 (p. 504) y = 3x2 ln x −7x2 9.4.31 (p. 504) y = 3(4x2 + 9) 2 + x 2 −ex 2 + e−x 2 + e2x 4 | Elementary Differential Equations with Boundary Value Problems_Page_766_Chunk3178 |
Answers to Selected Exercises 757 9.4.32 (p. 504) y = x ln x + x −√x + 1 x + 1 √x. 9.4.33 (p. 504) y = x3 ln |x| + x −2x3 + 1 x −1 x2 9.4.35 (p. 505) yp = Z x x0 e(x−t) −3e−(x−t) + 2e−2(x−t) 6 F(t) dt 9.4.36 (p. 505) yp = Z x x0 (x −t)2(2x + t) 6xt3 F(t) dt 9.4.37 (p. 505) yp = Z x x0 xe(x−t) −x2 + x(t −1) t4 F(t) dt 9.4.38 (p. 505) yp = Z x x0 x2 −t(t −2) −2te(x−t) 2x(t −1)2 F(t) dt 9.4.39 (p. 505) yp = Z x x0 e2(x−t) −2e(x−t) + 2e−(x−t) −e−2(x−t) 12 F(t) dt 9.4.40 (p. 505) yp = Z x x0 (x −t)3 6x F(t) dt 9.4.41 (p. 505) yp = Z x x0 (x + t)(x −t)3 12x2t3 F(t) dt 9.4.42 (p. 505) yp = Z x x0 e2(x−t)(1 + 2t) + e−2(x−t)(1 −2t) −4x2 + 4t2 −2 32t2 F(t) dt Section 10.1 Answers, pp. 514–515 10.1.1 (p. 514) Q′ 1 = 2 −1 10 Q1 + 1 25Q2 Q′ 2 = 6 + 3 50 Q1 −1 20Q2. 10.1.2 (p. 514) Q′ 1 = 12 − 5 100 + 2tQ1 + 1 100 + 3tQ2 Q′ 2 = 5 + 1 50 + tQ1 − 4 100 + 3tQ2. 10.1.3 (p. 514) m1y′′ 1 = −(c1 + c2)y′ 1 + c2y′ 2 −(k1 + k2)y1 + k2y2 + F1 m2y′′ 2 = (c2 −c3)y′ 1 −(c2 + c3)y′ 2 + c3y′ 3 + (k2 −k3)y1 −(k2 + k3)y2 + k3y3 + F2 m3y′′ 3 = c3y′ 1 + c3y′ 2 −c3y′ 3 + k3y1 + k3y2 −k3y3 + F3 10.1.4 (p. 515) x′′ = −α mx′ + gR2x (x2 + y2 + z2)3/2 y′′ = −α my′ + gR2y (x2 + y2 + z2)3/2 z′′ = −α mz′ + gR2z (x2 + y2 + z2)3/2 10.1.5 (p. 515) (a) x′ 1 = x2 x′ 2 = x3 x′ 3 = f(t, x1, y1, y2) y′ 1 = y2 y′ 2 = g(t, y1, y2) (b) u′ 1 = f(t,u1, v1, v2, w2) ; v′ 1 = v2 v′ 2 = g(t, u1, v1, v2, w1) w′ 1 = w2 w′ 2 = h(t, u1, v1, v2, w1, w2) (c) y′ 1 = y2 y′ 2 = y3 y′ 3 = f(t, y1, y2, y3) (d) y′ 1 = y2 y′ 2 = y3 y′ 3 = y4 y′ 4 = f(t, y1) (e) x′ 1 = x2 x′ 2 = f(t,x1, y1) y′ 1 = y2 y′ 2 = g(t, x1, y1) 10.1.6 (p. 515) x′ = x1 y′ = y1 z′ = z1 x′ 1 = − gR2x (x2 + y2 + z2)3/2 y′ 1 = − gR2y (x2 + y2 + z2)3/2 z′ 1 = − gR2z (x2 + y2 + z2)3/2 Section 10.2 Answers, pp. 518–521 | Elementary Differential Equations with Boundary Value Problems_Page_767_Chunk3179 |
758 Answers to Selected Exercises 10.2.1 (p. 518) (a) y′ = 2 4 4 2 y (b) y′ = −2 −2 −5 1 y (c) y′ = −4 −10 3 7 y (d) y′ = 2 1 1 2 y 10.2.2 (p. 518) (a) y′ = " −1 2 3 0 1 6 0 0 −2 # y (b) y′ = " 0 2 2 2 0 2 2 2 0 # y (c) y′ = " −1 2 2 2 −1 2 2 2 −1 # y (d) y′ = " 3 −1 −1 −2 3 2 4 −1 −2 # y 10.2.3 (p. 518) (a) y′ = 1 1 −2 4 y, y(0) = 1 0 (b) y′ = 5 3 −1 1 y, y(0) = 9 −5 10.2.4 (p. 519) (a) y′ = " 6 4 4 −7 −2 −1 7 4 3 # y, y(0) = " 3 −6 4 # (b) y′ = " 8 7 7 −5 −6 −9 5 7 10 # y, y(0) = " 2 −4 3 # 10.2.5 (p. 519) (a) y′ = −3 2 −5 3 + 3 −2t 6 −3t (b) y′ = 3 1 −1 1 y + −5et et 10.2.10 (p. 521) (a) d dtY 2 = Y ′Y + Y Y ′ (b) d dtY n = Y ′Y n−1 + Y Y ′Y n−2 + Y 2Y ′Y n−3 + · · · + Y n−1Y ′ = n−1 X r=0 Y rY ′Y n−r−1 10.2.13 (p. 521) B = (P ′ + PA)P −1. Section 10.3 Answers, pp. 525–529 10.3.2 (p. 525) y′ = 0 1 −P2(x) P0(x) −P1(x) P0(x) y 10.3.3 (p. 526) y′ = 0 1 · · · 0 ... ... ... ... 0 0 · · · 1 −Pn(x) P0(x) −Pn−1(x) P0(x) · · · −P1(x) P0(x) y 10.3.7 (p. 527) (b) y = 3e6t −6e−2t 3e6t + 6e−2t (c) y = 1 2 e6t + e−2t e6t −e−2t e6t −e−2t e6t + e−2t k 10.3.8 (p. 528) (b) y = 6e−4t + 4e3t 6e−4t −10e3t (c) y = 1 7 5e−4t + 2e3t 2e−4t −2e3t 5e−4t −5e3t 2e−4t + 5e3t k 10.3.9 (p. 528) (b) y = −15e2t −4et 9e2t + 2et (c) y = −5e2t + 6et −10e2t + 10et 3e2t −3et 6e2t −5et k 10.3.10 (p. 528) (b) y = 5e3t −3et 5e3t + 3et (c) y = 1 2 e3t + et e3t −et e3t −et e3t + et k 10.3.11 (p. 528) (b) y = " e2t −2e3t + 3e−t 2e3t −9e−t e2t −2e3t + 21e−t # (c) y = 1 6 " 4e2t + 3e3t −e−t 6e2t −6e3t 2e2t −3e3t + e−t −3e3t + 3e−t 6e3t 3e3t −3e−t 4e2t + 3e3t −7e−t 6e2t −6e3t 2e2t −3e3t + 7e−t # k | Elementary Differential Equations with Boundary Value Problems_Page_768_Chunk3180 |
Answers to Selected Exercises 759 10.3.12 (p. 528) (b) y = 1 3 " −e−2t + e4t −10e−2t + e4t 11e−2t + e4t # (c) y = 1 3 " 2e−2t + e4t −e−2t + e4t −e−2t + e4t −e−2t + e4t 2e−2t + e4t −e−2t + e4t −e−2t + e4t −e−2t + e4t 2e−2t + e4t # k 10.3.13 (p. 528) (b) y = " 3et + 3e−t −e−2t 3et + 2e−2t −e−2t # (c) y = " e−t et −e−t 2et −3e−t + e−2t 0 et 2et −2e−2t 0 0 e−2t # k 10.3.14 (p. 529) Y Z−1 and ZY −1 Section 10.4 Answers, pp. 539–541 10.4.1 (p. 539) y = c1 1 1 e3t + c2 1 −1 e−t 10.4.2 (p. 539) y = c1 1 1 e−t/2 + c2 −1 1 e−2t 10.4.3 (p. 539) y = c1 −3 1 e−t + c2 −1 2 e−2t 10.4.4 (p. 539) y = c1 2 1 e−3t + c2 −2 1 et 10.4.5 (p. 539) y = c1 1 1 e−2t + c1 −4 1 e3t 10.4.6 (p. 539) y = c1 3 2 e2t + c2 1 1 et 10.4.7 (p. 539) y = c1 −3 1 e−5t + c2 −1 1 e−3t 10.4.8 (p. 539) y = c1 " 1 2 1 # e−3t + c2 " −1 −4 1 # e−t + c3 " −1 −1 1 # e2t 10.4.9 (p. 539) y = c1 " 2 1 2 # e−16t + c2 " −1 2 0 # e2t + c3 " −1 0 1 # e2t 10.4.10 (p. 539) y = c1 " −2 −4 3 # et + c2 " −1 1 0 # e−2t + c3 " −7 −5 4 # e2t 10.4.11 (p. 539) y = c1 " −1 −1 1 # e−2t + c2 " −1 −2 1 # e−3t + c3 " −2 −6 3 # e−5t 10.4.12 (p. 539) y = c1 " 11 7 1 # e3t + c2 " 1 2 1 # e−2t + c3 " 1 1 1 # e−t 10.4.13 (p. 539) y = c1 " 4 −1 1 # e−4t + c2 " −1 −1 1 # e6t + c3 " −1 0 1 # e4t 10.4.14 (p. 539) y = c1 " 1 1 5 # e−5t + c2 " −1 0 1 # e5t + c3 " 1 1 0 # e5t 10.4.15 (p. 539) y = c1 " 1 −1 2 # + c2 " −1 0 3 # e6t + c3 " 1 3 0 # e6t 10.4.16 (p. 540) y = − 2 6 e5t + 4 2 e−5t 10.4.17 (p. 540) y = 2 −4 et/2 + −2 1 et 10.4.18 (p. 540) y = 7 7 e9t − 2 4 e−3t 10.4.19 (p. 540) y = 3 9 e5t − 4 2 e−5t | Elementary Differential Equations with Boundary Value Problems_Page_769_Chunk3181 |
760 Answers to Selected Exercises 10.4.20 (p. 540) y = " 5 5 0 # et/2 + " 0 0 1 # et/2 + " −1 2 0 # e−t/2 10.4.21 (p. 540) y = " 3 3 3 # et + " −2 −2 2 # e−t 10.4.22 (p. 540) y = " 2 −2 2 # et − " 3 0 3 # e−2t + " 1 1 0 # e3t 10.4.23 (p. 540) y = − " 1 2 1 # et + " 4 2 4 # e−t + " 1 1 0 # e2t 10.4.24 (p. 540) y = " −2 −2 2 # e2t − " 0 3 0 # e−2t + " 4 12 4 # e4t 10.4.25 (p. 540) y = " −1 −1 1 # e−6t + " 2 −2 2 # e2t + " 7 −7 −7 # e4t 10.4.26 (p. 540) y = " 1 4 4 # e−t + " 6 6 −2 # e2t 10.4.27 (p. 540) y = " 4 −2 2 # + " 3 −9 6 # e4t + " −1 1 −1 # e2t 10.4.29 (p. 541) Half lines of L1 : y2 = y1 and L2 : y2 = −y1 are trajectories other trajectories are asymptotically tangent to L1 as t →−∞and asymptotically tangent to L2 as t →∞. 10.4.30 (p. 541) Half lines of L1 : y2 = −2y1 and L2 : y2 = −y1/3 are trajectories other trajectories are asymptotically parallel to L1 as t →−∞and asymptotically tangent to L2 as t →∞. 10.4.31 (p. 541) Half lines of L1 : y2 = y1/3 and L2 : y2 = −y1 are trajectories other trajectories are asymptotically tangent to L1 as t →−∞and asymptotically parallel to L2 as t →∞. 10.4.32 (p. 541) Half lines of L1 : y2 = y1/2 and L2 : y2 = −y1 are trajectories other trajectories are asymptotically tangent to L1 as t →−∞and asymptotically tangent to L2 as t →∞. 10.4.33 (p. 541) Half lines of L1 : y2 = −y1/4 and L2 : y2 = −y1 are trajectories other trajectories are asymptotically tangent to L1 as t →−∞and asymptotically parallel to L2 as t →∞. 10.4.34 (p. 541) Half lines of L1 : y2 = −y1 and L2 : y2 = 3y1 are trajectories other trajectories are asymptotically parallel to L1 as t →−∞and asymptotically tangent to L2 as t →∞. 10.4.36 (p. 541) Points on L2 : y2 = y1 are trajectories of constant solutions. The trajectories of nonconstant solutions are half-lines on either side of L1, parallel to 1 −1 , traversed toward L1. 10.4.37 (p. 541) Points on L1 : y2 = −y1/3 are trajectories of constant solutions. The trajectories of nonconstant solutions are half-lines on either side of L1, parallel to −1 2 , traversed away from L1. 10.4.38 (p. 541) Points on L1 : y2 = y1/3 are trajectories of constant solutions. The trajectories of nonconstant solutions are half-lines on either side of L1, parallel to 1 −1 , 1 −1, traversed away from L1. | Elementary Differential Equations with Boundary Value Problems_Page_770_Chunk3182 |
Answers to Selected Exercises 761 10.4.39 (p. 541) Points on L1 : y2 = y1/2 are trajectories of constant solutions. The trajectories of nonconstant solutions are half-lines on either side of L1, parallel to 1 −1 , L1. 10.4.40 (p. 541) Points on L2 : y2 = −y1 are trajectories of constant solutions. The trajectories of nonconstant solutions are half-lines on either side of L2, parallel to −4 1 , traversed toward L1. 10.4.41 (p. 541) Points on L1 : y2 = 3y1 are trajectories of constant solutions. The trajectories of nonconstant solutions are half-lines on either side of L1, parallel to 1 −1 , traversed away from L1. Section 10.5 Answers, pp. 554–556 10.5.1 (p. 554) y = c1 2 1 e5t + c2 −1 0 e5t + 2 1 te5t . 10.5.2 (p. 554) y = c1 1 1 e−t + c2 1 0 e−t + 1 1 te−t 10.5.3 (p. 554) y = c1 −2 1 e−9t + c2 −1 0 e−9t + −2 1 te−9t 10.5.4 (p. 554) y = c1 −1 1 e2t + c2 −1 0 e2t + −1 1 te2t 10.5.5 (p. 554) c1 −2 1 + c2 −1 0 e−2t 3 + −2 1 te−2t 10.5.6 (p. 554) y = c1 3 2 e−4t + c2 −1 0 e−4t 2 + 3 2 te−4t 10.5.7 (p. 554) y = c1 4 3 e−t + c2 −1 0 e−t 3 + 4 3 te−t 10.5.8 (p. 554) y = c1 " −1 −1 2 # + c2 " 1 1 2 # e4t + c3 " 0 1 0 # e4t 2 + " 1 1 2 # te4t ! 10.5.9 (p. 554) y = c1 " −1 1 1 # et + c2 " 1 −1 1 # e−t + c3 " 0 3 0 # e−t + " 1 −1 1 # te−t ! . 10.5.10 (p. 554) y = c1 " 0 1 1 # e2t + c2 " 1 0 1 # e−2t + c3 " 1 1 0 # e−2t 2 + " 1 0 1 # te−2t ! 10.5.11 (p. 554) y = c1 " −2 −3 1 # e2t + c2 " 0 −1 1 # e4t + c3 " 1 0 0 # e4t 2 + " 0 −1 1 # te4t ! 10.5.12 (p. 554) y = c1 " −1 −1 1 # e−2t + c2 " 1 1 1 # e4t + c3 " 1 0 0 # e4t 2 + " 1 1 1 # te4t ! . 10.5.13 (p. 554) y = 6 2 e−7t − 8 4 te−7t 10.5.14 (p. 554) y = 5 8 e3t − 12 16 te3t | Elementary Differential Equations with Boundary Value Problems_Page_771_Chunk3183 |
762 Answers to Selected Exercises 10.5.15 (p. 554) y = 2 3 e−5t − 8 4 te−5t 10.5.16 (p. 554) y = 3 1 e5t − 12 6 te5t 10.5.17 (p. 554) y = 0 2 e−4t + 6 6 te−4t 10.5.18 (p. 554) y = " 4 8 −6 # et + " 2 −3 −1 # e−2t + " −1 1 0 # te−2t 10.5.19 (p. 554) y = " 3 3 6 # e2t − " 9 5 6 # + " 2 2 0 # t 10.5.20 (p. 554) y = − " 2 0 2 # e−3t + " −4 9 1 # et − " 0 4 4 # tet 10.5.21 (p. 555) y = " −2 2 2 # e4t + " 0 −1 1 # e2t + " 3 −3 3 # te2t 10.5.22 (p. 555) y = − " 1 1 0 # e−4t + " −3 2 −3 # e8t + " 8 0 −8 # te8t 10.5.23 (p. 555) y = " 3 6 3 # e4t − " 3 4 1 # + " 8 4 4 # t 10.5.24 (p. 555) y = c1 " 0 1 1 # e6t + c2 " −1 1 0 # e6t 4 + " 0 1 1 # te6t ! +c3 " 1 1 0 # e6t 8 + " −1 1 0 # te6t 4 + " 0 1 1 # t2e6t 2 ! 10.5.25 (p. 555) y = c1 " −1 1 1 # e3t + c2 " 1 0 0 # e3t 2 + " −1 1 1 # te3t ! +c3 " 1 2 0 # e3t 36 + " 1 0 0 # te3t 2 + " −1 1 1 # t2e3t 2 ! 10.5.26 (p. 555) y = c1 " 0 −1 1 # e−2t + c2 " −1 1 0 # e−2t + " 0 −1 1 # te−2t ! +c3 " 3 −2 0 # e−2t 4 + " −1 1 0 # te−2t + " 0 −1 1 # t2e−2t 2 ! | Elementary Differential Equations with Boundary Value Problems_Page_772_Chunk3184 |
Answers to Selected Exercises 763 10.5.27 (p. 555) y = c1 " 0 1 1 # e2t + c2 " 1 1 0 # e2t 2 + " 0 1 1 # te2t ! +c3 " −1 1 0 # e2t 8 + " 1 1 0 # te2t 2 + " 0 1 1 # t2e2t 2 ! 10.5.28 (p. 555) y = c1 " −2 1 2 # e−6t + c2 − " 6 1 0 # e−6t 6 + " −2 1 2 # te−6t ! +c3 − " 12 1 0 # e−6t 36 − " 6 1 0 # te−6t 6 + " −2 1 2 # t2e−6t 2 ! . 10.5.29 (p. 555) y = c1 " −4 0 1 # e−3t + c2 " 6 1 0 # e−3t + c3 " 1 0 0 # e−3t + " 2 1 1 # te−3t ! 10.5.30 (p. 555) y = c1 " −1 0 1 # e−3t + c2 " 0 1 0 # e−3t + c3 " 1 0 0 # e−3t + " −1 −1 1 # te−3t ! 10.5.31 (p. 555) y = c1 " 2 0 1 # e−t + c2 " −3 2 0 # e−t + c3 " 1 0 0 # e−t 2 + " −1 2 1 # te−t ! 10.5.32 (p. 555) y = c1 " −1 1 0 # e−2t + c2 " 0 0 1 # e−2t + c3 " −1 0 0 # e−2t + " 1 −1 1 # te−2t ! Section 10.6 Answers, pp. 565–568 10.6.1 (p. 565) y = c1e2t 3 cos t + sin t 5 cos t + c2e2t 3 sin t −cos t 5 sin t . 10.6.2 (p. 565) y = c1e−t 5 cos 2t + sin 2t 13 cos 2t + c2e−t 5 sin 2t −cos 2t 13 sin 2t . 10.6.3 (p. 565) y = c1e3t cos 2t + sin 2t 2 cos 2t + c2e3t sin 2t −cos 2t 2 sin 2t . 10.6.4 (p. 565) y = c1e2t cos 3t −sin 3t cos 3t + c2e2t sin 3t + cos 3t sin 3t . 10.6.5 (p. 566) y = c1 " −1 −1 2 # e−2t + c2e4t " cos 2t −sin 2t cos 2t + sin 2t 2 cos 2t # + c3e4t " sin 2t + cos 2t sin 2t −cos 2t 2 sin 2t # . 10.6.6 (p. 566) y = c1 " −1 −1 1 # e−t + c2e−2t " cos 2t −sin 2t −cos 2t −sin 2t 2 cos 2t # + c3e−2t " sin 2t + cos 2t −sin 2t + cos 2t 2 sin 2t # 10.6.7 (p. 566) y = c1 " 1 1 1 # e2t + c2et " −sin t sin t cos t # + c3et " cos t −cos t sin t # | Elementary Differential Equations with Boundary Value Problems_Page_773_Chunk3185 |
764 Answers to Selected Exercises 10.6.8 (p. 566) y = c1 " −1 1 1 # et + c2e−t " −sin 2t −cos 2t 2 cos 2t 2 cos 2t # + c3e−t " cos 2t −sin 2t 2 sin 2t 2 sin 2t # 10.6.9 (p. 566) y = c1e3t cos 6t −3 sin 6t 5 cos 6t + c2e3t sin 6t + 3 cos 6t 5 sin 6t 10.6.10 (p. 566) y = c1e2t cos t −3 sin t 2 cos t + c2e2t sin t + 3 cos t 2 sin t 10.6.11 (p. 566) y = c1e2t 3 sin 3t −cos 3t 5 cos 3t + c2e2t −3 cos 3t −sin 3t 5 sin 3t 10.6.12 (p. 566) y = c1e2t sin 4t −8 cos 4t 5 cos 4t + c2e2t −cos 4t −8 sin 4t 5 sin 4t 10.6.13 (p. 566) y = c1 " −1 1 1 # e−2t + c2et " sin t −cos t cos t # + c3et " −cos t −sin t sin t # 10.6.14 (p. 566) y = c1 " 2 2 1 # e−2t + c2e2t " −cos 3t −sin 3t −sin 3t cos 3t # + c3e2t " −sin 3t + cos 3t cos 3t sin 3t # 10.6.15 (p. 566) y = c1 " 1 2 1 # e3t + c2e6t " −sin 3t sin 3t cos 3t # + c3e6t " cos 3t −cos 3t sin 3t # 10.6.16 (p. 566) y = c1 " 1 1 1 # et + c2et " 2 cos t −2 sin t cos t −sin t 2 cos t # + c3et " 2 sin t + 2 cos t cos t + sin t 2 sin t # 10.6.17 (p. 566) y = et 5 cos 3t + sin 3t 2 cos 3t + 3 sin 3t 10.6.18 (p. 566) y = e4t 5 cos 6t + 5 sin 6t cos 6t −3 sin 6t 10.6.19 (p. 566) y = et 17 cos 3t −sin 3t 7 cos 3t + 3 sin 3t 10.6.20 (p. 566) y = et/2 cos(t/2) + sin(t/2) −cos(t/2) + 2 sin(t/2) 10.6.21 (p. 566) y = " 1 −1 2 # et + e4t " 3 cos t + sin t cos t −3 sin t 4 cos t −2 sin t # 10.6.22 (p. 566) y = " 4 4 2 # e8t + e2t " 4 cos 2t + 8 sin 2t −6 sin 2t + 2 cos 2t 3 cos 2t + sin 2t # 10.6.23 (p. 566) y = " 0 3 3 # e−4t + e4t " 15 cos 6t + 10 sin 6t 14 cos 6t −8 sin 6t 7 cos 6t −4 sin 6t # 10.6.24 (p. 566) y = " 6 −3 3 # e8t + " 10 cos 4t −4 sin 4t 17 cos 4t −sin 4t 3 cos 4t −7 sin 4t # 10.6.29 (p. 567) U = 1 √ 2 −1 1 , V = 1 √ 2 1 1 10.6.30 (p. 567) U ≈ .5257 .8507 , V ≈ −.8507 .5257 | Elementary Differential Equations with Boundary Value Problems_Page_774_Chunk3186 |
Answers to Selected Exercises 765 10.6.31 (p. 567) U ≈ .8507 .5257 , V ≈ −.5257 .8507 10.6.32 (p. 567) U ≈ −.9732 .2298 , V ≈ .2298 .9732 10.6.33 (p. 567) U ≈ .5257 .8507 , V ≈ −.8507 .5257 10.6.34 (p. 567) U ≈ −.5257 .8507 , V ≈ .8507 .5257 10.6.35 (p. 568) U ≈ −.8817 .4719 , V ≈ .4719 .8817 10.6.36 (p. 568) U ≈ .8817 .4719 , V ≈ −.4719 .8817 10.6.37 (p. 568) U = 0 1 , V = −1 0 10.6.38 (p. 568) U = 0 1 , V = 1 0 10.6.39 (p. 568) U = 1 √ 2 1 1 , V = 1 √ 2 −1 1 10.6.40 (p. 568) U ≈ .5257 .8507 , V ≈ −.8507 .5257 Section 10.7 Answers, pp. 575–577 10.7.1 (p. 575) 5e4t + e−3t(2 + 8t) −e4t −e−3t(1 −4t) 10.7.2 (p. 575) 13e3t + 3e−3t −e3t −11e−3t 10.7.3 (p. 575) 1 9 7 −6t −11 + 3t 10.7.4 (p. 575) 5 −3et −6 + 5et 10.7.5 (p. 575) e−5t(3 + 6t) + e−3t(3 −2t) −e−5t(3 + 2t) −e−3t(1 −2t) 10.7.6 (p. 575) t 0 10.7.7 (p. 575) −1 6 " 2 −6t 7 + 6t 1 −12t # 10.7.8 (p. 575) −1 6 " 3et + 4 6et −4 10 # 10.7.9 (p. 575) 1 18 " et(1 + 12t) −e−5t(1 + 6t) −2et(1 −6t) −e−5t(1 −12t) et(1 + 12t) −e−5t(1 + 6t) # 10.7.10 (p. 575) 1 3 " 2et et 2et # 10.7.11 (p. 575) t sin t 0 10.7.12 (p. 575) − t2 2t 10.7.13 (p. 575) (t −1) (ln |t −1| + t) 1 −1 10.7.14 (p. 575) 1 9 5e2t −e−3t e3t −5e−2t 10.7.15 (p. 576) 1 4t 2t3 ln |t| + t3(t + 2) 2 ln |t| + 3t −2 10.7.16 (p. 576) 1 2 te−t(t + 2) + (t3 −2) tet(t −2) + (t3 + 2) 10.7.17 (p. 576) − " t t t # 10.7.18 (p. 576) 1 4 " −3et 1 e−t # 10.7.19 (p. 576) " 2t2 + t t −t # 10.7.20 (p. 576) et 4t " 2t + 1 2t −1 2t + 1 # 10.7.22 (p. 576) (a) y′ = 0 1 · · · 0 0 0 · · · 0 ... ... ... ... 0 0 · · · 1 −Pn(t)/P0(t) −Pn−1/P0(t) · · · −P1(t)/P0(t) y + 0 0 ... F(t)/P0(t) . | Elementary Differential Equations with Boundary Value Problems_Page_775_Chunk3187 |
766 Answers to Selected Exercises (b) y1 y2 · · · yn y′ 1 y′ 2 · · · y′ n ... ... ... ... y(n−1) 1 y(n−1) 2 · · · y(n−1) n Section 11.1 Answers, pp. 585–586 11.1.2 (p. 585) λn = n2, yn = sin nx, n = 1, 2, 3, . . . 11.1.3 (p. 585) λ0 = 0, y0 = 1; λn = n2, yn = cos nx, n = 1, 2, 3, . . . 11.1.4 (p. 585) λn = (2n −1)2 4 , yn = sin (2n −1)x 2 , n = 1, 2, 3, . . . , 11.1.5 (p. 585) λn = (2n −1)2 4 , yn = cos (2n −1)x 2 , n = 1, 2, 3, . . . 11.1.6 (p. 585) λ0 = 0, y0 = 1, λn = n2, y1n = cos nx, y2n = sin nx, n = 1, 2, 3, . . . 11.1.7 (p. 585) λn = n2π2, yn = cos nπx, n = 1, 2, 3, . . . 11.1.8 (p. 585) λn = (2n −1)2π2 4 , yn = cos (2n −1)πx 2 , n = 1, 2, 3, . . . 11.1.9 (p. 585) λn = n2π2, yn = sin nπx, n = 1, 2, 3, . . . 11.1.10 (p. 585) λ0 = 0, y0 = 1, λn = n2π2, y1n = cos nπx, y2n = sin nπx, n = 1, 2, 3, . . . 11.1.11 (p. 585) λn = (2n −1)2π2 4 , yn = sin (2n −1)πx 2 , n = 1, 2, 3, . . . 11.1.12 (p. 585) λ0 = 0, y0 = 1, λn = n2π2 4 , y1n = cos nπx 2 , y2n = sin nπx 2 , n = 1, 2, 3, . . . 11.1.13 (p. 585) λn = n2π2 4 , yn = sin nπx 2 , n = 1, 2, 3, . . . 11.1.14 (p. 585) λn = (2n −1)2π2 36 , yn = cos (2n −1)πx 6 , n = 1, 2, 3, . . . 11.1.15 (p. 585) λn = (2n −1)2π2, yn = sin(2n −1)πx, n = 1, 2, 3, . . . 11.1.16 (p. 585) λn = n2π2 25 , yn = cos nπx 5 , n = 1, 2, 3, . . . 11.1.23 (p. 586) λn = 4n2π2/L2 yn = sin 2nπx L , n = 1, 2, 3, . . . 11.1.24 (p. 586) λn = n2π2/L2 yn = cos nπx L , n = 1, 2, 3, . . . 11.1.25 (p. 586) λn = 4n2π2/L2 yn = sin 2nπx L , n = 1, 2, 3, . . . 11.1.26 (p. 586) λn = n2π2/L2 yn = cos nπx L , n = 1, 2, 3, . . . . Section 11.2 Answers, pp. 598–602 11.2.2 (p. 598) F(x) = 2 + 2 π ∞ X n=1 (−1)n n sin nπx; F(x) = ( 2, x = −1, 2 −x, −1 < x < 1, 2, x = 1 11.2.3 (p. 598) F(x) = −π2 −12 ∞ X n=1 (−1)n n2 cos nx −4 ∞ X n=1 (−1)n n sin nx; F(x) = ( −3π2, x = −π, 2x −3x2, −π < x < π, −3π2, x = π 11.2.4 (p. 598) F(x) = −12 π2 ∞ X n=1 (−1)n cos nπx n2 ; F(x) = 1 −3x2 −1 ≤x ≤1 | Elementary Differential Equations with Boundary Value Problems_Page_776_Chunk3188 |
Answers to Selected Exercises 767 11.2.5 (p. 598) F(x) = 2 π −4 π ∞ X n=1 1 4n2 −1 cos 2nx; F(x) = | sin x|, −π ≤x ≤π 11.2.6 (p. 599) F(x) = −1 2 sin x + 2 ∞ X n=2 (−1)n n n2 −1 sin nx;; F(x) = x cos x, −π ≤x ≤π 11.2.7 (p. 599) F(x) = −2 π + π 2 cos x −4 π ∞ X n=1 4n2 + 1 (4n2 −1)2 cos 2nx; F(x) = |x|cos x, −π ≤x ≤π 11.2.8 (p. 599) F(x) = 1 −1 2 cos x −2 ∞ X n=2 (−1)n n2 −1 cos nx; F(x) = x sin x, −π ≤x ≤π 11.2.9 (p. 599) F(x) = π 2 sin x −16 π ∞ X n=1 n (4n2 −1)2 sin 2nx; F(x) = |x|sin x, −π ≤x ≤π 11.2.10 (p. 599) F(x) = 1 π + 1 2 cos πx −2 π ∞ X n=1 (−1)n 4n2 −1 cos 2nπx; F(x) = f(x), −1 ≤x ≤1 11.2.11 (p. 599) F(x) = 1 4π sin πx −8 π2 ∞ X n=1 (−1)n n (4n2 −1)2 sin 2nπx; −1 4π ∞ X n=1 (−1)n n(n + 1) sin(2n + 1)πx F(x) = f(x), −1 ≤x ≤1 11.2.12 (p. 599) F(x) = 1 2 sin πx−4 π ∞ X n=1 (−1)n n 4n2 −1 sin 2nπx; F(x) = 0, −1 ≤x < 1 2, −1 2 , x = −1 2 , sin πx, −1 2 < x < 1 2, 1 2 , x = 1 2, 0, 1 2 < x ≤1 11.2.13 (p. 599) F(x) = 1 π + 1 π cos πx −2 π ∞ X n=2 1 n2 −1 1 −n sin nπ 2 cos nπx; F(x) = 0, −1 ≤x < 1 2 , 1 2 , x = −1, | sin πx|, −1 2 < x < 1 2 , 1 2 , x = 1, 0, 1 2 < x ≤1 11.2.14 (p. 599) F(x) = 1 π2 + 1 4π cos πx + 2 π2 ∞ X n=1 (−1)n 4n2 + 1 (4n2 −1)2 cos 2nπx + 1 4π ∞ X n=1 (−1)n 2n + 1 n(n + 1) cos(2n + 1)πx; F(x) = 0, −1 ≤x < 1 2, 1 4, x = −1 2, x sin πx, −1 2 < x < 1 2, 1 4, x = 1 2, 0, 1 2 < x ≤1, 11.2.15 (p. 599) F(x) = 1 −8 π2 ∞ X n=0 1 (2n + 1)2 cos (2n + 1)πx 4 −4 π ∞ X n=1 (−1)n n sin nπx 4 ; | Elementary Differential Equations with Boundary Value Problems_Page_777_Chunk3189 |
768 Answers to Selected Exercises F(x) = 2, x = −4, 0, −4 < x < 0, x, 0 ≤x < 4, 2, x = 4 11.2.16 (p. 599) F(x) = 1 2 + 1 π ∞ X n=1 1 n sin 2nπx + 8 π3 ∞ X n=0 1 (2n + 1)3 sin(2n + 1)πx; F(x) = 1 2 , x = −1, x2, −1 < x < 0, 1 2 , x = 0, 1 −x2, 0 < x < 1, 1 2 , x = 1 11.2.17 (p. 599) F(x) = 3 4 + 1 π ∞ X n=1 1 n sin nπ 2 cos nπx 2 + 3 π ∞ X n=1 1 n cos nπ −cos nπ 2 sin nπx 2 11.2.18 (p. 599) F(x) = 5 2 + 3 π ∞ X n=1 1 n sin 2nπ 3 cos nπx 3 + 1 π ∞ X n=1 1 n cos nπ −cos 2nπ 3 sin nπx 3 11.2.20 (p. 599) F(x) = sinh π π 1 + 2 ∞ X n=1 (−1)n n2 + 1 cos nx −2 ∞ X n=1 (−1)nn n2 + 1 sin nx ! 11.2.21 (p. 599) F(x) = −π cos x −1 2 sin x + 2 ∞ X n=2 (−1)n n n2 −1 sin nx 11.2.22 (p. 600) F(x) = 1 −1 2 cos x −π sin x −2 ∞ X n=2 (−1)n n2 −1 cos nx 11.2.23 (p. 600) F(x) = −2 sin kπ π ∞ X n=1 (−1)n n n2 −k2 sin nx 11.2.24 (p. 600) F(x) = sin kπ π " 1 k −2k ∞ X n=1 (−1)n n2 −k2 cos nx # Section 11.3 Answers, pp. 613–616 11.3.1 (p. 613) C(x) = L2 3 + 4L2 π2 ∞ X n=1 (−1)n n2 cos nπx L 11.3.2 (p. 613) C(x) = 1 2 + 4 π2 ∞ X n=1 1 (2n −1)2 cos(2n −1)πx 11.3.3 (p. 613) C(x) = −2L2 3 + 4L2 π2 ∞ X n=1 1 n2 cos nπx L 11.3.4 (p. 613) C(x) = 1 −cos kπ kπ −2k π ∞ X n=1 [1 −(−1)n cos kπ] n2 −k2 cos nx. 11.3.5 (p. 613) C(x) = 1 2 −2 π ∞ X n=1 (−1)n 2n −1 cos (2n −1)πx L 11.3.6 (p. 613) C(x) = −2L2 3 + 4L2 π2 ∞ X n=1 (−1)n n2 cos nπx L | Elementary Differential Equations with Boundary Value Problems_Page_778_Chunk3190 |
Answers to Selected Exercises 769 11.3.7 (p. 613) C(x) = 1 3 + 4 π2 ∞ X n=1 1 n2 cos nπx 11.3.8 (p. 613) C(x) = eπ −1 π + 2 π ∞ X n=1 [(−1)neπ −1] (n2 + 1) cos nx 11.3.9 (p. 613) C(x) = L2 6 −L2 π2 ∞ X n=1 1 n2 cos 2nπx L 11.3.10 (p. 613) C(x) = −2L2 3 + 4L2 π2 ∞ X n=1 1 n2 cos nπx L 11.3.11 (p. 614) S(x) = 4 π ∞ X n=1 1 (2n −1) sin (2n −1)πx L 11.3.12 (p. 614) S(x) = 2 π ∞ X n=1 1 n sin nπx 11.3.13 (p. 614) S(x) = 2 π ∞ X n=1 [1 −(−1)n cos kπ] n n2 −k2 sin nx 11.3.14 (p. 614) S(x) = 2 π ∞ X n=1 1 n h 1 −cos nπ 2 i sin nπx L 11.3.15 (p. 614) S(x) = 4L π2 ∞ X n=1 (−1)n+1 (2n −1)2 sin (2n −1)πx L 11.3.16 (p. 614) S(x) = π 2 sin x −16 π ∞ X n=1 n (4n2 −1)2 sin 2nx 11.3.17 (p. 614) S(x) = −2 π ∞ X n=1 n[(−1)neπ −1] (n2 + 1) sin nx 11.3.18 (p. 614) CM(x) = −4 π ∞ X n=1 (−1)n 2n −1 cos (2n −1)πx 2L 11.3.19 (p. 614) CM(x) = −4L2 π ∞ X n=1 (−1)n 2n −1 1 − 8 (2n −1)2π2 cos (2n −1)πx 2L 11.3.20 (p. 614) CM(x) = −4 π ∞ X n=1 (−1)n + 2 (2n −1)π cos (2n −1)πx 2 . 11.3.21 (p. 614) CM(x) = −4 π ∞ X n=1 1 2n −1 cos (2n + 1)π 4 cos (2n −1)πx 2L 11.3.22 (p. 614) CM(x) = 4 π ∞ X n=1 (−1)n 2n −1 (2n −3)(2n + 1) cos (2n −1)x 2 11.3.23 (p. 614) CM(x) = −8 π ∞ X n=1 1 (2n −3)(2n + 1) cos (2n −1)x 2 11.3.24 (p. 614) CM(x) = −8L2 π2 ∞ X n=1 1 (2n −1)2 1 + 4(−1)n (2n −1)π cos (2n −1)πx 2L | Elementary Differential Equations with Boundary Value Problems_Page_779_Chunk3191 |
770 Answers to Selected Exercises 11.3.25 (p. 614) SM(x) = 4 π ∞ X n=1 1 (2n −1) sin (2n −1)πx 2L 11.3.26 (p. 614) SM(x) = −16L2 π2 ∞ X n=1 1 (2n −1)2 (−1)n + 2 (2n −1)π sin (2n −1)πx 2L 11.3.27 (p. 614) SM(x) = 4 π ∞ X n=1 1 2n −1 1 −cos (2n −1)π) 4 sin (2n −1)πx 2L 11.3.28 (p. 614) SM(x) = 4 π ∞ X n=1 2n −1 (2n −3)(2n + 1) sin (2n −1)x 2 11.3.29 (p. 614) SM(x) = 8 π ∞ X n=1 (−1)n (2n −3)(2n + 1) sin (2n −1)x 2 11.3.30 (p. 614) SM(x) = 8L2 π2 ∞ X n=1 1 (2n −1)2 (−1)n + 4 (2n −1)π sin (2n −1)πx 2L 11.3.31 (p. 614) C(x) = −7L4 5 −144L4 π4 ∞ X n=1 (−1)n n4 cos nπx L 11.3.32 (p. 614) C(x) = −2L4 5 −48L4 π4 ∞ X n=1 1 + (−1)n2 n4 cos nπx L 11.3.33 (p. 614) C(x) = 3L4 5 −48L4 π4 ∞ X n=1 2 + (−1)n n4 cos nπx L 11.3.34 (p. 614) C(x) = L4 30 −3L4 π4 ∞ X n=1 1 n4 cos 2nπx L 11.3.36 (p. 615) S(x) = 8L2 π3 ∞ X n=1 1 (2n −1)3 sin (2n −1)πx L 11.3.37 (p. 615) S(x) = −4L3 π3 ∞ X n=1 (1 + (−1)n2) n3 sin nπx L 11.3.38 (p. 615) S(x) = −12L3 π3 ∞ X n=1 (−1)n n3 sin nπx L 11.3.39 (p. 615) S(x) = 96L4 π5 ∞ X n=1 1 (2n −1)5 sin (2n −1)πx L 11.3.40 (p. 615) S(x) = −720L5 π5 ∞ X n=1 (−1)n n5 sin nπx L 11.3.41 (p. 615) S(x) = −240L5 π5 ∞ X n=1 1 + (−1)n2 n5 sin nπx L 11.3.43 (p. 615) CM(x) = −64L3 π3 ∞ X n=1 1 (2n −1)3 (−1)n + 3 (2n −1)π cos (2n −1)πx 2L 11.3.44 (p. 615) CM(x) = −32L2 π3 ∞ X n=1 (−1)n (2n −1)3 cos (2n −1)πx 2L | Elementary Differential Equations with Boundary Value Problems_Page_780_Chunk3192 |
Answers to Selected Exercises 771 11.3.45 (p. 615) CM(x) = −96L3 π3 ∞ X n=1 1 (2n −1)3 (−1)n + 2 (2n −1)π cos (2n −1)πx 2L 11.3.46 (p. 615) CM(x) = 96L3 π3 ∞ X n=1 1 (2n −1)3 (−1)n3 + 4 (2n −1)π cos (2n −1)πx 2L 11.3.47 (p. 615) CM(x) = 96L3 π3 ∞ X n=1 1 (2n −1)3 (−1)n5 + 8 (2n −1)π cos (2n −1)πx 2L 11.3.48 (p. 615) CM(x) = −384L4 π4 ∞ X n=1 1 (2n −1)4 1 + (−1)n4 (2n −1)π cos (2n −1)πx 2L 11.3.49 (p. 615) CM(x) = −768L4 π4 ∞ X n=1 1 (2n −1)4 1 + (−1)n2 (2n −1)π cos (2n −1)πx 2L 11.3.51 (p. 615) SM(x) = 32L2 π3 ∞ X n=1 1 (2n −1)3 sin (2n −1)πx 2L 11.3.52 (p. 615) SM(x) = −96L3 π3 ∞ X n=1 1 (2n −1)3 1 + (−1)n 4 (2n −1)π sin (2n −1)πx 2L 11.3.53 (p. 616) SM(x) = 96L3 π3 ∞ X n=1 1 (2n −1)3 1 + (−1)n 2 (2n −1)π sin (2n −1)πx 2L 11.3.54 (p. 616) SM(x) = 192L3 π4 ∞ X n=1 (−1)n (2n −1)4 sin (2n −1)πx 2L 11.3.55 (p. 616) SM(x) = 1536L4 π4 ∞ X n=1 1 (2n −1)4 (−1)n + 3 (2n −1)π sin (2n −1)πx 2L 11.3.56 (p. 616) SM(x) = 384L4 π4 ∞ X n=1 1 (2n −1)4 (−1)n + 4 (2n −1)π sin (2n −1)πx 2L Section 12.1 Answers, pp. 626–629 12.1.8 (p. 626) u(x,t) = 8 π3 ∞ X n=1 1 (2n −1)3 e−(2n−1)2π2t sin(2n −1)πx 12.1.9 (p. 626) u(x,t) = 4 π ∞ X n=1 1 (2n −1)e−9(2n−1)2π2t/16 sin (2n −1)πx 4 12.1.10 (p. 626) u(x,t) = π 2 e−3t sin x −16 π ∞ X n=1 n (4n2 −1)2 e−12n2t sin 2nx 12.1.11 (p. 626) u(x,t) = −32 π3 ∞ X n=1 (1 + (−1)n2) n3 e−9n2π2t/4 sin nπx 2 12.1.12 (p. 626) u(x,t) = −324 π3 ∞ X n=1 (−1)n n3 e−4n2π2t/9 sin nπx 3 12.1.13 (p. 626) u(x,t) = 8 π2 ∞ X n=1 (−1)n+1 (2n −1)2 e−(2n−1)2π2t sin (2n −1)πx 2 12.1.14 (p. 626) u(x,t) = −720 π5 ∞ X n=1 (−1)n n5 e−7n2π2t sin nπx | Elementary Differential Equations with Boundary Value Problems_Page_781_Chunk3193 |
772 Answers to Selected Exercises 12.1.15 (p. 626) u(x,t) = 96 π5 ∞ X n=1 1 (2n −1)5 e−5(2n−1)2π2t sin(2n −1)πx 12.1.16 (p. 626) u(x,t) = −240 π5 ∞ X n=1 1 + (−1)n2 n5 e−2n2π2t sin nπx. 12.1.17 (p. 627) u(x,t) = 16 3 + 64 π2 ∞ X n=1 (−1)n n2 e−9π2n2t/16 cos nπx 4 12.1.18 (p. 627) u(x,t) = −8 3 + 16 π2 ∞ X n=1 1 n2 e−n2π2t cos nπx 2 12.1.19 (p. 627) u(x,t) = 1 6 −1 π2 ∞ X n=1 1 n2 e−36n2π2t cos 2nπx 12.1.20 (p. 627) u(x,t) = 4 −384 π4 ∞ X n=1 1 (2n −1)4 e−3(2n−1)2π2t/4 cos (2n −1)πx 2 12.1.21 (p. 627) u(x,y) = −28 5 −576 π4 ∞ X n=1 (−1)n n4 e−5n2π2t/2 cos nπx √ 2 12.1.22 (p. 627) u(x,t) = −2 5 −48 π4 ∞ X n=1 1 + (−1)n2 n4 e−3n2π2t cos nπx 12.1.23 (p. 627) u(x,t) = 3 5 −48 π4 ∞ X n=1 2 + (−1)n n4 e−n2π2t cos nπx 12.1.24 (p. 627) u(x,t) = π4 30 −3 ∞ X n=1 1 n4 e−4n2t cos 2nx 12.1.25 (p. 627) u(x,t) = 8 π ∞ X n=1 (−1)n (2n + 1)(2n −3)e−(2n−1)2π2t/4 sin (2n −1)πx 2 12.1.26 (p. 627) u(x,t) = 8 ∞ X n=1 1 (2n −1)2 (−1)n + 4 (2n −1)π e−3(2n−1)2t/4 sin (2n −1)x 2 12.1.27 (p. 627) u(x,t) = 128 π3 ∞ X n=1 1 (2n −1)3 e−5(2n−1)2t/16 sin (2n −1)πx 4 12.1.28 (p. 627) u(x,t) = −96 π3 ∞ X n=1 1 (2n −1)3 1 + (−1)n 4 (2n −1)π e−(2n−1)2π2t/4 sin (2n −1)πx 2 12.1.29 (p. 627) u(x,t) = 96 π3 ∞ X n=1 1 (2n −1)3 1 + (−1)n 2 (2n −1)π e−(2n−1)2π2t/4 sin (2n −1)πx 2 12.1.30 (p. 627) u(x,t) = 192 π4 ∞ X n=1 (−1)n (2n −1)4 e−(2n−1)2π2t/4 sin (2n −1)πx 2 12.1.31 (p. 627) u(x,t) = 1536 π4 ∞ X n=1 1 (2n −1)4 (−1)n + 3 (2n −1)π e−(2n−1)2π2t/4 sin (2n −1)πx 2 12.1.32 (p. 628) u(x,t) = 384 π4 ∞ X n=1 1 (2n −1)4 (−1)n + 4 (2n −1)π e−(2n−1)2π2t/4 sin (2n −1)πx 2 | Elementary Differential Equations with Boundary Value Problems_Page_782_Chunk3194 |
Answers to Selected Exercises 773 12.1.33 (p. 628) u(x,t) = −64 ∞ X n=1 e−3(2n−1)2t/4 (2n −1)3 (−1)n + 3 (2n −1)π cos (2n −1)x 2 12.1.34 (p. 628) u(x,t) = −16 π ∞ X n=1 (−1)n 2n −1e−(2n−1)2t cos (2n −1)x 4 12.1.35 (p. 628) u(x,t) = −64 π ∞ X n=1 (−1)n 2n −1 1 − 8 (2n −1)2π2 e−9(2n−1)2π2t/64 cos (2n −1)πx 8 12.1.36 (p. 628) u(x,t) = 8 π2 ∞ X n=1 1 (2n −1)2 e−3(2n−1)2π2t/4 cos (2n −1)πx 2 12.1.37 (p. 628) u(x,t) = −96 π3 ∞ X n=1 1 (2n −1)3 (−1)n + 2 (2n −1)π e−(2n−1)2π2t/4 cos (2n −1)πx 2 12.1.38 (p. 628) u(x,t) = −32 π ∞ X n=1 (−1)n (2n −1)3 e−7(2n−1)2t/4 cos (2n −1)x 2 12.1.39 (p. 628) u(x,t) = 96 π3 ∞ X n=1 1 (2n −1)3 (−1)n5 + 8 (2n −1)π e−(2n−1)2π2t/4 cos (2n −1)πx 2 12.1.40 (p. 628) u(x,t) = 96 π3 ∞ X n=1 1 (2n −1)3 (−1)n3 + 4 (2n −1)π e−(2n−1)2π2t/4 cos (2n −1)πx 2 12.1.41 (p. 628) u(x,t) = −768 π4 ∞ X n=1 1 (2n −1)4 1 + (−1)n2 (2n −1)π e−(2n−1)2π2t/4 cos (2n −1)πx 2 12.1.42 (p. 628) u(x,t) = −384 π4 ∞ X n=1 1 (2n −1)4 1 + (−1)n4 (2n −1)π e−(2n−1)2π2t/4 cos (2n −1)πx 2 12.1.43 (p. 628) u(x,t) = 1 2 −2 π ∞ X n=1 (−1)n 2n −1e−(2n−1)2π2a2t/L2 cos (2n −1)πx L 12.1.44 (p. 628) u(x,t) = 2 π ∞ X n=1 1 n h 1 −cos nπ 2 i e−n2π2a2t/L2 sin nπx L 12.1.45 (p. 629) u(x,t) = 4 π ∞ X n=1 1 2n −1 sin (2n −1)π 4 e−(2n−1)2π2a2t/4L2 cos (2n −1)πx 2L 12.1.46 (p. 629) u(x,t) = 4 π ∞ X n=1 1 2n −1 1 −cos (2n −1)π) 4 e−(2n−1)2π2a2t/4L2 sin (2n −1)πx 2L 12.1.48 (p. 629) u(x,t) = 1 −x + x3 + 4 π ∞ X n=1 e−9π2(2n−1)2t/16 (2n −1) sin (2n −1)πx 4 12.1.49 (p. 629) u(x,t) = 1 + x + x2 −8 π3 ∞ X n=1 e−(2n−1)2π2t (2n −1)3 sin(2n −1)πx 12.1.50 (p. 629) u(x,t) = −1 −x + x3 + 8 π2 ∞ X n=1 1 (2n −1)2 e−3(2n−1)2π2t/4 cos (2n −1)πx 2 12.1.51 (p. 629) u(x, t) = x2 −x −2 −64 π ∞ X n=1 (−1)n 2n −1 1 − 8 (2n −1)2π2 e−9(2n−1)2π2t/64 cos (2n −1)πx 8 | Elementary Differential Equations with Boundary Value Problems_Page_783_Chunk3195 |
774 Answers to Selected Exercises 12.1.52 (p. 629) u(x,t) = sin πx + 8 π ∞ X n=1 (−1)n (2n + 1)(2n −3)e−(2n−1)2π2t/4 sin (2n −1)πx 2 12.1.53 (p. 629) u(x,t) = x3 −x + 3 + 32 π3 ∞ X n=1 e−(2n−1)2π2t/4 (2n −1)3 sin (2n −1)πx 2 Section 12.2 Answers, pp. 642–649 12.2.1 (p. 642) u(x,t) = 4 3π3 ∞ X n=1 (−1)n+1 (2n −1)3 sin 3(2n −1)πt sin(2n −1)πx 12.2.2 (p. 642) u(x,t) = 8 π3 ∞ X n=1 1 (2n −1)3 cos 3(2n −1)πt sin(2n −1)πx 12.2.3 (p. 642) u(x,t) = −4 π3 ∞ X n=1 (1 + (−1)n2) n3 cos n √ 7 πt sin nπx 12.2.4 (p. 642) u(x,t) = 8 3π4 ∞ X n=1 1 (2n −1)4 sin 3(2n −1)πt sin(2n −1)πx 12.2.5 (p. 642) u(x,t) = − 4 √ 7 π4 ∞ X n=1 (1 + (−1)n2) n4 sin n √ 7 πt sin nπx 12.2.6 (p. 642) u(x,t) = 324 π3 ∞ X n=1 (−1)n n3 cos 8nπt 3 sin nπx 3 12.2.7 (p. 642) u(x,t) = 96 π5 ∞ X n=1 1 (2n −1)5 cos 2(2n −1)πt sin(2n −1)πx 12.2.8 (p. 642) u(x,t) = 243 2π4 ∞ X n=1 (−1)n n4 sin 8nπt 3 sin nπx 3 12.2.9 (p. 642) u(x,t) = 48 π6 ∞ X n=1 1 (2n −1)6 sin 2(2n −1)πt sin(2n −1)πx. 12.2.10 (p. 642) u(x,t) = π 2 cos √ 5 t sin x −16 π ∞ X n=1 n (4n2 −1)2 cos 2n √ 5 t sin 2nx 12.2.11 (p. 642) u(x,t) = −240 π5 ∞ X n=1 1 + (−1)n2 n5 cos nπt sin nπx 12.2.12 (p. 642) u(x,t) = π 2 √ 5 sin √ 5 t sin x − 8 π √ 5 ∞ X n=1 1 (4n2 −1)2 sin 2n √ 5 t sin 2nx 12.2.13 (p. 642) u(x,t) = −240 π6 ∞ X n=1 1 + (−1)n2 n6 sin nπt sin nπx 12.2.14 (p. 643) u(x,t) = −720 π5 ∞ X n=1 (−1)n n5 cos 3nπt sin nπx 12.2.15 (p. 643) u(x,t) = −240 π6 ∞ X n=1 (−1)n n6 sin 3nπt sin nπx 12.2.18 (p. 644) u(x,t) = −128 π3 ∞ X n=1 (−1)n (2n −1)3 cos 3(2n −1)πt 4 cos (2n −1)πx 4 | Elementary Differential Equations with Boundary Value Problems_Page_784_Chunk3196 |
Answers to Selected Exercises 775 12.2.19 (p. 644) u(x,t) = −64 π3 ∞ X n=1 1 (2n −1)3 (−1)n + 3 (2n −1)π cos(2n −1)πt cos (2n −1)πx 2 12.2.20 (p. 644) u(x,t) = −512 3π4 ∞ X n=1 (−1)n (2n −1)4 sin 3(2n −1)πt 4 cos (2n −1)πx 4 12.2.21 (p. 644) u(x,t) = −64 π4 ∞ X n=1 1 (2n −1)4 (−1)n + 3 (2n −1)π sin(2n −1)πt cos (2n −1)πx 2 12.2.22 (p. 644) u(x,t) = 96 π3 ∞ X n=1 1 (2n −1)3 (−1)n3 + 4 (2n −1)π cos (2n −1) √ 5 πt 2 cos (2n −1)πx 2 12.2.23 (p. 644) u(x,t) = −96 ∞ X n=1 1 (2n −1)3 (−1)n + 2 (2n −1)π cos (2n −1) √ 3 t 2 cos (2n −1)x 2 12.2.24 (p. 644) u(x,t) = 192 π4√ 5 ∞ X n=1 1 (2n −1)4 (−1)n3 + 4 (2n −1)π sin (2n −1) √ 5 πt 2 cos (2n −1)πx 2 12.2.25 (p. 644) u(x,t) = −192 √ 3 ∞ X n=1 1 (2n −1)4 (−1)n + 2 (2n −1)π sin (2n −1) √ 3t 2 sin (2n −1)x 2 12.2.26 (p. 644) u(x,t) = −384 π4 ∞ X n=1 1 (2n −1)4 1 + (−1)n4 (2n −1)π cos 3(2n −1)πt 2 cos (2n −1)πx 2 12.2.27 (p. 644) u(x,t) = 96 π3 ∞ X n=1 1 (2n −1)3 (−1)n5 + 8 (2n −1)π cos (2n −1) √ 7 πt 2 cos (2n −1)πx 2 12.2.28 (p. 644) u(x,t) = −768 3π5 ∞ X n=1 1 (2n −1)5 1 + (−1)n4 (2n −1)π sin 3(2n −1)πt 2 cos (2n −1)πx 2 12.2.29 (p. 644) u(x,t) = 192 π4√ 7 ∞ X n=1 1 (2n −1)4 (−1)n5 + 8 (2n −1)π sin (2n −1) √ 7 πt 2 cos (2n −1)πx 2 12.2.30 (p. 645) u(x,t) = −768 π4 ∞ X n=1 1 (2n −1)4 1 + (−1)n2 (2n −1)π cos (2n −1)πt 2 cos (2n −1)πx 2 12.2.31 (p. 645) u(x,t) = −1536 π5 ∞ X n=1 1 (2n −1)5 1 + (−1)n2 (2n −1)π sin (2n −1)πt 2 cos (2n −1)πx 2 12.2.32 (p. 645) u(x,t) = 1 2 [CMf(x + at) + CMf(x −at)] + 1 2a Z x+at x−at CMg(τ) dτ 12.2.35 (p. 645) u(x,t) = 32 π ∞ X n=1 1 (2n −1)3 cos 4(2n −1)t sin (2n −1)x 2 12.2.36 (p. 645) u(x,t) = −96 π3 ∞ X n=1 1 (2n −1)3 1 + (−1)n 4 (2n −1)π cos 3(2n −1)πt 2 sin (2n −1)πx 2 12.2.37 (p. 645) u(x,t) = 8 π ∞ X n=1 1 (2n −1)4 sin 4(2n −1)t sin (2n −1)x 2 12.2.38 (p. 646) u(x,t) = −64 π4 ∞ X n=1 1 (2n −1)4 1 + (−1)n 4 (2n −1)π sin 3(2n −1)πt 2 sin (2n −1)πx 2 | Elementary Differential Equations with Boundary Value Problems_Page_785_Chunk3197 |
776 Answers to Selected Exercises 12.2.39 (p. 646) u(x,t) = 96 π3 ∞ X n=1 1 (2n −1)3 1 + (−1)n 2 (2n −1)π cos 3(2n −1)πt 2 sin (2n −1)πx 2 12.2.40 (p. 646) u(x,t) = 192 π ∞ X n=1 (−1)n (2n −1)4 cos (2n −1) √ 3 t 2 sin (2n −1)x 2 12.2.41 (p. 646) u(x,t) = 64 π4 ∞ X n=1 1 (2n −1)4 1 + (−1)n 2 (2n −1)π sin 3(2n −1)πt 2 sin (2n −1)πx 2 12.2.42 (p. 646) u(x,t) = 384 √ 3 π ∞ X n=1 (−1)n (2n −1)5 sin (2n −1) √ 3 t 2 sin (2n −1)x 2 12.2.43 (p. 646) u(x,t) = 1536 π4 ∞ X n=1 1 (2n −1)4 (−1)n + 3 (2n −1)π cos (2n −1) √ 5 πt 2 sin (2n −1)πx 2 12.2.44 (p. 646) u(x,t) = 384 π4 ∞ X n=1 1 (2n −1)4 (−1)n + 4 (2n −1)π cos(2n −1)πt sin (2n −1)πx 2 12.2.45 (p. 646) u(x,t) = 3072 √ 5 π5 ∞ X n=1 1 (2n −1)5 (−1)n + 3 (2n −1)π sin (2n −1) √ 5 πt 2 sin (2n −1)πx 2 12.2.46 (p. 646) u(x,t) = 384 π5 ∞ X n=1 1 (2n −1)5 (−1)n + 4 (2n −1)π sin(2n −1)πt sin (2n −1)πx 2 12.2.47 (p. 646) u(x,t) = 1 2 [SMf(x + at) + SMf(x −at)] + 1 2a Z x+at x−at SMg(τ) dτ 12.2.50 (p. 647) u(x,t) = 4 −768 π4 ∞ X n=1 1 (2n −1)4 cos √ 5(2n −1)πt 2 cos (2n −1)πx 2 12.2.51 (p. 647) u(x,t) = 4t −1536 √ 5 π5 ∞ X n=1 1 (2n −1)5 sin √ 5(2n −1)πt 2 cos (2n −1)πx 2 12.2.52 (p. 647) u(x,t) = −2π4 5 −48 ∞ X n=1 1 + (−1)n2 n4 cos 2nt cos nx 12.2.53 (p. 647) u(x,t) = −7 5 −144 π4 ∞ X n=1 (−1)n n4 cos n √ 7 πt cos nπx 12.2.54 (p. 647) u(x,t) = −2π4t 5 −24 ∞ X n=1 1 + (−1)n2 n5 sin 2nt cos nx 12.2.55 (p. 647) u(x,t) = −7t 5 − 144 π5√ 7 ∞ X n=1 (−1)n n5 sin n √ 7 πt cos nπx 12.2.56 (p. 647) u(x,t) = π4 30 −3 ∞ X n=1 1 n4 cos 8nt cos 2nx 12.2.57 (p. 647) u(x,t) = 3 5 −48 π4 ∞ X n=1 2 + (−1)n n4 cos nπt cos nπx 12.2.58 (p. 647) u(x,t) = π4t 30 −3 8 ∞ X n=1 1 n5 sin 8nt cos 2nx | Elementary Differential Equations with Boundary Value Problems_Page_786_Chunk3198 |
Answers to Selected Exercises 777 12.2.59 (p. 647) u(x,t) = 3t 5 −48 π5 ∞ X n=1 2 + (−1)n n5 sin nπt cos nπx 12.2.60 (p. 647) u(x,t) = 1 2 [Cf(x + at) + Cf (x −at)] + 1 2a R x+at x−at Cg(τ) dτ 12.2.63 (p. 648) (c) u(x, t) = f(x + at) + f(x −at) 2 + 1 2a Z x+at x−at g(u) du 12.2.64 (p. 649) u(x,t) = x(1 + 4at) 12.2.65 (p. 649) u(x,t) = x2 + a2t2 + t 12.2.66 (p. 649) u(x,t) = sin(x + at) 12.2.67 (p. 649) u(x, t) = x3 + 6tx2 + 3a2t2x + 2a2t3 12.2.68 (p. 649) u(x,t) = x sin x cos at + at cos x sin at + sin x sin at a Section 12.3 Answers, pp. 662–665 12.3.1 (p. 662) u(x,y) = 8 π3 ∞ X n=1 sinh(2n −1)π(1 −y) (2n −1)3 sinh(2n −1)π sin(2n −1)πx 12.3.2 (p. 662) u(x,y) = −32 π3 ∞ X n=1 (1 + (−1)n2) sinh nπ(3 −y)/2 n3 sinh 3nπ/2 sin nπx 2 12.3.3 (p. 662) u(x,y) = 8 π2 ∞ X n=1 (−1)n+1 sinh(2n −1)π(1 −y/2) (2n −1)2 sinh(2n −1)π sin (2n −1)πx 2 12.3.4 (p. 662) u(x,y) = π 2 sinh(1 −y) sinh 1 sin x −16 π ∞ X n=1 n sinh 2n(1 −y) (4n2 −1)2 sinh 2n sin 2nx 12.3.5 (p. 662) u(x,y) = 3y + 108 π3 ∞ X n=1 (−1)n sinh nπy/3 n3 cosh 2nπ/3 cos nπx 3 12.3.6 (p. 662) u(x,y) = y 2 + 4 π3 ∞ X n=1 sinh(2n −1)πy (2n −1)3 cosh 2(2n −1)π cos(2n −1)πx 12.3.7 (p. 662) u(x,y) = −8y 3 + 32 π3 ∞ X n=1 (−1)n sinh nπy/2 n3 cosh nπ cos nπx 2 12.3.8 (p. 662) u(x,y) = y 3 + 4 π3 ∞ X n=1 sinh nπy n3 cosh nπ cos nπx 12.3.9 (p. 662) u(x,y) = 128 π3 ∞ X n=1 cosh(2n −1)π(x −3)/4 (2n −1)3 cosh 3(2n −1)π/4 sin (2n −1)πy 4 12.3.10 (p. 662) u(x,y) = −96 π3 ∞ X n=1 1 + (−1)n 4 (2n −1)π cosh(2n −1)π(x −2)/2 (2n −1)3 cosh(2n −1)π sin (2n −1)πy 2 12.3.11 (p. 662) u(x,y) = 768 π3 ∞ X n=1 1 + (−1)n 2 (2n −1)π cosh(2n −1)π(x −2)/4 (2n −1)3 cosh(2n −1)π/2 sin (2n −1)πy 4 12.3.12 (p. 662) u(x,y) = 96 π3 ∞ X n=1 3 + (−1)n 4 (2n −1)π cosh(2n −1)π(x −3)/2 (2n −1)3 cosh 3(2n −1)π/2 sin (2n −1)πy 2 12.3.13 (p. 663) u(x,y) = −16 π ∞ X n=1 cosh(2n −1)x/2 (2n −3)(2n + 1)(2n −1) sinh(2n −1)/2 cos (2n −1)y 2 12.3.14 (p. 663) u(x,y) = −432 π3 ∞ X n=1 1 + 4(−1)n (2n −1)π cosh(2n −1)πx/6 (2n −1)3 sinh(2n −1)π/3 cos (2n −1)πy 6 | Elementary Differential Equations with Boundary Value Problems_Page_787_Chunk3199 |
778 Answers to Selected Exercises 12.3.15 (p. 663) u(x,y) = −64 π ∞ X n=1 (−1)n cosh(2n −1)x/2 (2n −1)4 sinh(2n −1)/2 cos (2n −1)y 2 . 12.3.16 (p. 663) u(x,y) = −192 π4 ∞ X n=1 cosh(2n −1)πx/2 (2n −1)4 sinh(2n −1)π/2 (−1)n + 2 (2n −1)π cos (2n −1)πy 2 12.3.17 (p. 663) u(x,y) = ∞ X n=1 αn sinh nπy/a sinh nπb/a sin nπx a , αn = 2 a Z a 0 f(x) sin nπx a dx u(x, y) = 72 π3 ∞ X n=1 sinh(2n −1)πy/3 (2n −1)3 sinh 2(2n −1)π/3 sin (2n −1)πx 3 12.3.18 (p. 663) u(x,y) = α0(1 −y/b) + ∞ X n=1 αn sinh nπ(b −y)/a sinh nπb/a cos nπx a , α0 = 1 a Z a 0 f(x) dx, αn = 2 a Z a 0 f(x)cos nπx a dx, n ≥1 u(x, y) = 8(1 −y) 15 −48 π4 ∞ X n=1 1 n4 sinh nπ(1 −y) sinh nπ cos nπx 12.3.19 (p. 663) u(x,y) = ∞ X n=1 αn sinh(2n −1)π(b −y)/2a sinh(2n −1)πb/2a cos (2n −1)πx 2a , αn = 2 a Z a 0 f(x)cos (2n −1)πx 2a dx u(x, y) = 288 π3 ∞ X n=1 sinh(2n −1)π(2 −y)/6 (2n −1)3 sinh(2n −1)π/3 sin (2n −1)πx 6 12.3.20 (p. 663) u(x,y) = ∞ X n=1 αn sinh(2n −1)π(b −y)/2a sinh(2n −1)πb/2a sin (2n −1)πx 2a , αn = 2 a Z a 0 f(x)sin (2n −1)πx 2a dx u(x, y) = 32 π3 ∞ X n=1 (−1)n5 + 18 (2n −1)π sinh(2n −1)π(2 −y)/2 (2n −1)3 sinh(2n −1)π cos (2n −1)πx 2 . 12.3.21 (p. 663) u(x,y) = ∞ X n=1 αn cosh nπ(y −b)/a cosh nπb/a sin nπx a , αn = 2 a Z a 0 f(x) sin nπx a dx u(x, y) = −12 ∞ X n=1 (−1)n cosh n(y −2) n3 cosh 2n sin nx 12.3.22 (p. 663) u(x,y) = α0 + ∞ X n=1 αn cosh nπy/a cosh nπb/a cos nπx a , α0 = 1 a Z a 0 f(x) dx, αn = 2 a Z a 0 f(x)cos nπx a dx, n ≥1 u(x, y) = π4 30 −3 ∞ X n=1 1 n4 cosh 2ny cos 2n cos 2nx 12.3.23 (p. 663) u(x,y) = a π ∞ X n=1 αn sinh nπ(y −b)/a n cosh nπb/a sin nπx a , αn = 2 a Z a 0 f(x) sin nπx a dx u(x, y) = 4 π ∞ X n=1 (−1)n+1 sinh(2n −1)(y −1) (2n −1)3 cosh(2n −1) sin(2n −1)x | Elementary Differential Equations with Boundary Value Problems_Page_788_Chunk3200 |
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