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# Sect5 1

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### Sect5 1

1. 1. SECTION 5.1 INTRODUCTION: SECOND-ORDER LINEAR EQUATIONS In this section the central ideas of the theory of linear differential equations are introduced and illustrated concretely in the context of second-order equations. These key concepts include superposition of solutions (Theorem 1), existence and uniqueness of solutions (Theorem 2), linear independence, the Wronskian (Theorem 3), and general solutions (Theorem 4). This discussion of second-order equations serves as preparation for the treatment of nth order linear equations in Section 5.2. Although the concepts in this section may seem somewhat abstract to students, the problems set is quite tangible and largely computational. In each of Problems 1-16 the verification that y1 and y2 satisfy the given differential equation is a routine matter. As in Example 2, we then impose the given initial conditions on the general solution y = c1y1 + c2y2. This yields two linear equations that determine the values of the constants c1 and c2. 1. Imposition of the initial conditions y (0) = 0, y ′(0) = 5 on the general solution y ( x ) = c1e x + c2e − x yields the two equations c1 + c2 = 0, c1 − c2 = 0 with solution c1 = 5 / 2, c2 = − 5 / 2. Hence the desired particular solution is y(x) = 5(ex - e-x)/2. 2. Imposition of the initial conditions y (0) = − 1, y′(0) = 15 on the general solution y ( x) = c1e3 x + c2 e −3 x yields the two equations c1 + c2 = − 1, 3c1 − 3c2 = 15 with solution c1 = 2, c2 = 3. Hence the desired particular solution is y(x) = 2e3x - 3e-3x. 3. Imposition of the initial conditions y (0) = 3, y′(0) = 8 on the general solution y ( x ) = c1 cos 2 x + c2 sin 2 x yields the two equations c1 = 3, 2c2 = 8 with solution c1 = 3, c2 = 4. Hence the desired particular solution is y(x) = 3 cos 2x + 4 sin 2x. 4. Imposition of the initial conditions y (0) = 10, y′(0) = − 10 on the general solution y ( x) = c1 cos 5 x + c2 sin 5 x yields the two equations c1 = 10, 5c2 = − 10 with solution c1 = 3, c2 = 4. Hence the desired particular solution is y(x) = 10 cos 5x - 2 sin 5x. 5. Imposition of the initial conditions y (0) = 1, y′(0) = 0 on the general solution y ( x ) = c1e x + c2e 2 x yields the two equations c1 + c2 = 1, c1 + 2c2 = 0 with solution c1 = 2, c2 = − 1. Hence the desired particular solution is y(x) = 2ex - e2x. 6. Imposition of the initial conditions y (0) = 7, y′(0) = − 1 on the general solution y ( x) = c1e 2 x + c2e −3 x yields the two equations c1 + c2 = 7, 2c1 − 3c2 = − 1 with solution c1 = 4, c2 = 3. Hence the desired particular solution is y(x) = 4e2x + 3e-3x.
2. 2. 7. Imposition of the initial conditions y (0) = − 2, y′(0) = 8 on the general solution y ( x) = c1 + c2 e − x yields the two equations c1 + c2 = − 2, − c2 = 8 with solution c1 = 6, c2 = − 8. Hence the desired particular solution is y(x) = 6 − 8e-x. 8. Imposition of the initial conditions y (0) = 4, y′(0) = − 2 on the general solution y ( x) = c1 + c2e3 x yields the two equations c1 + c2 = 4, 3c2 = − 2 with solution c1 = 14 / 3, c2 = 2 / 3. Hence the desired particular solution is y(x) = (14 - 2e3x)/3. 9. Imposition of the initial conditions y (0) = 2, y′(0) = − 1 on the general solution y ( x ) = c1e − x + c2 x e − x yields the two equations c1 = 2, − c1 + c2 = − 1 with solution c1 = 2, c2 = 1. Hence the desired particular solution is y(x) = 2e-x + xe-x. 10. Imposition of the initial conditions y (0) = 3, y′(0) = 13 on the general solution y ( x ) = c1e5 x + c2 x e5 x yields the two equations c1 = 3, 5c1 + c2 = 13 with solution c1 = 3, c2 = − 2. Hence the desired particular solution is y(x) = 3e5x - 2xe5x. 11. Imposition of the initial conditions y (0) = 0, y ′(0) = 5 on the general solution y ( x ) = c1e x cos x + c2 e x sin x yields the two equations c1 = 0, c1 + c2 = 5 with solution c1 = 0, c2 = 5. Hence the desired particular solution is y(x) = 5ex sin x. 12. Imposition of the initial conditions y (0) = 2, y′(0) = 0 on the general solution y ( x) = c1e −3 x cos 2 x + c2e −3 x sin 2 x yields the two equations c1 = 2, − 3c1 + 2c2 = 5 with solution c1 = 2, c2 = 3. Hence the desired particular solution is y(x) = e-3x(2 cos 2x + 3 sin 2x). 13. Imposition of the initial conditions y (1) = 3, y ′(1) = 1 on the general solution y ( x) = c1 x + c2 x 2 yields the two equations c1 + c2 = 3, c1 + 2c2 = 1 with solution c1 = 5, c2 = − 2. Hence the desired particular solution is y(x) = 5x - 2x2. 14. Imposition of the initial conditions y (2) = 10, y ′(2) = 15 on the general solution y ( x) = c1 x 2 + c2 x −3 yields the two equations 4c1 + c2 / 8 = 10, 4c1 − 3c2 /16 = 15 with solution c1 = 3, c2 = − 16. Hence the desired particular solution is y(x) = 3x2 - 16/x3. 15. Imposition of the initial conditions y (1) = 7, y′(1) = 2 on the general solution y ( x) = c1 x + c2 x ln x yields the two equations c1 = 7, c1 + c2 = 2 with solution c1 = 7, c2 = − 5. Hence the desired particular solution is y(x) = 7x - 5x ln x. 16. Imposition of the initial conditions y (1) = 2, y′(1) = 3 on the general solution y ( x) = c1 cos(ln x) + c2 sin(ln x) yields the two equations c1 = 2, c2 = 3. Hence the desired particular solution is y(x) = 2 cos(ln x) + 3 sin(ln x).
3. 3. 17. If y = c / x then y′ + y 2 = − c / x 2 + c 2 / x 2 = c (c − 1) / x 2 ≠ 0 unless either c = 0 or c = 1. 18. If y = cx 3 then yy′′ = cx 3 ⋅ 6cx = 6c 2 x 4 ≠ 6 x 4 unless c 2 = 1. 19. If y = 1 + x then yy′′ + ( y ′)2 = (1 + x )(− x −3/ 2 / 4) + ( x −1/ 2 / 2) 2 = − x −3/ 2 / 4 ≠ 0. 20. Linearly dependent, because f(x) = π = π(cos2x + sin2x) = π g(x) Linearly independent, because x = + x x if x > 0, whereas x = − x x if x < 0. 3 2 3 2 21. 22. Linearly independent, because 1 + x = c(1 + x ) would require that c = 1 with x = 0, but c = 0 with x = -1. Thus there is no such constant c. 23. Linearly independent, because f(x) = +g(x) if x > 0, whereas f(x) = -g(x) if x < 0. 24. Linearly dependent, because g(x) = 2 f(x). 25. f(x) = ex sin x and g(x) = ex cos x are linearly independent, because f(x) = k g(x) would imply that sin x = k cos x, whereas sin x and cos x are linearly independent. 26. To see that f(x) and g(x) are linearly independent, assume that f(x) = c g(x), and then substitute both x = 0 and x = π/2. 27. Let L[y] = y″ + py′ + qy. Then L[yc] = 0 and L[yp] = f, so L[yc + yp] = L[yc] + [yp] = 0 + f = f. 28. If y(x) = 1 + c1 cos x + c2 sin x then y′(x) = -c1sin x + c2cos x, so the initial conditions y(0) = y′(0) = -1 yield c1 = -2, c2 = -1. Hence y = 1 - 2 cos x - sin x. 29. There is no contradiction because if the given differential equation is divided by x2 to get the form in Equation (8) in the text, then the resulting functions p(x) = -4/x and q(x) = 6/x2 are not continuous at x = 0. 30. (a) y2 = x 3 and y2 = x 3 are linearly independent because x 3 = c x 3 would require that c = 1 with x = 1, but c = -1 with x = -1. (b) The fact that W(y1, y2) = 0 everywhere does not contradict Theorem 3, because when the given equation is written in the required form
4. 4. y″ - (3/x)y′ + (3/x2)y = 0, the coefficient functions p(x) = -3/x and q(x) = 3/x2 are not continuous at x = 0. 31. W(y1, y2) = -2x vanishes at x = 0, whereas if y1 and y2 were (linearly independent) solutions of an equation y″ + py′ + qy = 0 with p and q both continuous on an open interval I containing x = 0, then Theorem 3 would imply that W ≠ 0 on I. 32. (a) W = y1y2′ - y1′y2, so AW' = A(y1′y2′ + y1y2″ - y1″y2 - y1′y2′) = y1(Ay2″) - y2(Ay1″) = y1(-By2′ - Cy2) - y2(-By1′ - Cy1) = -B(y1y2′ - y1′y2) and thus AW' = -BW. (b) Just separate the variables. (c) Because the exponential factor is never zero. In Problems 33-42 we give the characteristic equation, its roots, and the corresponding general solution. 33. r 2 − 3r + 2 = 0; r = 1, 2; y(x) = c1ex + c2e2x 34. r 2 + 2r − 15 = 0; r = 3, − 5; y(x) = c1e-5x +c2e3x 35. r 2 + 5r = 0; r = 0, − 5; y(x) = c1 + c2e-5x 36. 2r 2 + 3r = 0; r = 0, − 3 / 2; y(x) = c1 + c2e-3x/2 37. 2r 2 − r − 2 = 0; r = 1, − 1/ 2; y(x) = c1e-x/2 + c2ex 38. 4r 2 + 8r + 3 = 0; r = − 1/ 2, − 3 / 2; y(x) = c1e-x/2 +c2e-3x/2 39. 4r 2 + 4r + 1 = 0; r = − 1/ 2, − 1/ 2; y(x) = (c1 + c2x)e-x/2 40. 9r 2 − 12r + 4 = 0; r = − 2 / 3, − 2 / 3; y(x) = (c1 + c2x)e2x/3 41. 6r 2 − 7r − 20 = 0; r = − 4 / 3, 5 / 2; y(x) = c1e-4x/3 + c2e5x/2
5. 5. 42. 35r 2 − r − 12 = 0; r = − 4 / 7, 3 / 5; y(x) = c1e-4x/7 + c2e3x/5 In Problems 43-48 we first write and simplify the equation with the indicated characteristic roots, and then write the corresponding differential equation. 43. (r − 0)(r + 10) = r 2 + 10r = 0; y′′ + 10 y′ = 0 44. (r − 10)(r + 10) = r 2 − 100 = 0; y′′ − 100 y = 0 45. (r + 10)(r + 10) = r 2 + 20r + 100 = 0; y′′ + 20 y′ + 100 y = 0 46. (r − 10)(r − 100) = r 2 − 110r + 1000 = 0; y′′ − 110 y′ + 1000 y = 0 47. (r − 0)(r − 0) = r 2 = 0; y′′ = 0 48. (r − 1 − 2)(r − 1 + 2) = r 2 − 2r − 1 = 0; y′′ − 2 y′ − y = 0 −x −2 x 49. The solution curve with y( 0) = 1, y ′(0 ) = 6 is y( x ) = 8 e − 7 e . We find that y ′( x ) = 0 when x = ln(7/4) so e − x = 4 / 7 and e −2 x = 16 / 49 . It follows that y(ln(7 / 4)) = 16 / 7 , so the high point on the curve is (ln(7 / 4)), 16 / 7) ≈ (0.56, 2.29) , which looks consistent with Fig. 5.1.6. 50. The two solution curves with y(0) = a and y(0) = b (as well as y ′( 0) = 1) are y = (2 a + 1)e − x − ( a + 1)e −2 x , y = (2 b + 1)e − x − (b + 1)e −2 x . −x −2 x Subtraction and then division by a - b gives 2e = e , so it follows that x = -ln 2. Now substitution in either formula gives y = -2, so the common point of intersection is (-ln 2, -2).