Advanced Engineering Mathematics 8th Edition O’Neil SOLUTIONS 2024/2025. MANUAL. Chapter 2. Second-Order Differential Equations. LATEST 2024 UPDATE.

2.1 The Linear Second-Order Equation 1. It is a routine exercise in differentiation to show that y1 (x) and y2 (x) are solutions of the homogeneous equation, while yp (x) is a solution of the nonhomogeneous equation. The Wronskian of y1 (x) and y2 (x) is W (x) = sin(6x) cos(6x) = −6 sin2 (x) − 6 sin2 (x) = −6, 6 cos(6x) −6 sin(6x) and this is nonzero for all x, so these solutions are linearly independent on the real line. The general solution of the nonhomogeneous differential equation is 1 y = c1 sin(6x) + c2 cos(6x) + 36 (x − 1). For the initial value problem, we need y (0) = c2 − 36 1 = −5 so c2 = −179/36. And y0 (0) = 2 = 6c1 + so c1 = 71/216. The unique solution of the initial value problem is 71 216 sin(6x) − 179 y(x) = 36 1 cos(36 (x − 6x) 1). + 2. The Wronskian of e4x and e−4x is W (x) = e4x e−4x = −8 = 0 4e4x −4e−4x 37 so these solutions of the associated homogeneous equation are indepen- dent. With the particular solution yp (x) of the nonhomogeneous equation, this equation has general solution y(x) = c1 e4x + c2 e−4x − 14 x2 − 321 . From the initial conditions we obtain 1 y(0) = c1 + c2 − 32 = 12 and y0 (0) = 4c1 − 4c2 = 3. Solve these to obtain c1 = 409/64 and c2 = 361/64 to obtain the solution y(x) = 409 e4x + 361 e −4x − 1 x2 − 1 . 64 64 4 32 3. The associated homogeneous equation has solutions e−2x and e−x . Their Wronskian is W (x) = e−2x e−x −2e−2x −e−x = e−3x and this is nonzero for all x. The general solution of the nonhomogeneous differential equation is y(x) = c1 e−2x + c2 e−x + . For the initial value problem, solve 15 y(0) = −3 = c1 + c2 + 2 and y0 (0) = −1 = −2c1 − c2 to get c1 = 23/2, c2 = −22. The initial value problem has solution y(x) = 23 e−2x − 22e−x + 15 . 2 2 4. The associated homogeneous equation has solutions y1 (x) = e3x cos(2x), y2 (x) = e3x sin(2x). The Wronskian of these solutions is W (x) = e3x cos(2x) e3x sin(2x) = e6x = 0 3e3x cos(2x) − 2e3x sin(2x) 3e3x sin(2x) + 2e3x cos(2x) 2.1. LINEAR SECOND-ORDER EQUATION for all x. The general solution of the nonhomogeneous equation is y(x) = c1 e3x cos(2x) + c2 e3x sin(2x) − ex . To satisfy the initial conditions, it is required that 1 y(0) = −1 = c1 − 8 and 1 3c1 + 2c2 − = 1. 8 Solve these to obtain c1 = −7/8 and c2 = 15/8. The solution of the initial value problem is 7 3x e y(x) = − e cos(2x) + 158 3x sin( 2x) − 81 e x . 8 5. The associated homogeneous equation has solutions y1 (x) = ex cos(x), y2 (x) = ex sin(x). These have Wronskian W (x) = ex cos(x) ex sin(x) = e2x = 0 ex cos(x) − ex sin(x) ex sin(x) + ex cos(x) so these solutions are independent. The general solution of the nonhomo- geneous differential equation is y(x) = c1 ex cos(x) + c2 ex sin(x) − 5 2 2 5x − 52 . c − We need 5 y(0) = c1 − 2 = 6 and y0 (0) = 1 = c1 + c2 − 5. Solve these to get c1 = 17/2 and c2 = −5/2 to get the solution y(x) = 17 x cos(x) − 5 ex sin(x) − 5 x2 − 5x − 5 . e 2 2 2 2 6. Suppose y1 and y2 are solutions of the homogeneous equation (2.2). Then y001 + py10 + qy1 = 0 and y00 0 2 + py2 + qy2 = 0. Multiply the first equation by y2 and the second by −y1 and add the resulting equations to obtain y100 y2 − y 200 y1 + p(y 10 y2 − y 20 y1 ) = 0. We want to relate this equation to the Wronskian of these solutions, which is W = y1 y20 − y2 y 10 . Now W 0 = y1 y200 − y2y 100 . Then W 0 + pW = 0. This is a linear first-order differential equation for W . Multiply this equa- tion by the integrating factor R e p(x) dx to obtain R R W e p(x) dx + pW e p(x) dx = 0, which we can write as R 0 W e p(x) dx = 0. Integrate this to obtain R W e p(x) dx = k, with k constant. Then R W (x) = ke− p(x) dx . This shows that W (x) = 0 for all x (if k = 0), and W (x) = 0 for all x (if k = 0). Now suppose that y1 and y2 are independent and observe that d y2 y1 y0 −y 2 y0 1 = 2 y 2 1 = W (x). dx y1 1 y 12 If k = 0, then W (x) = 0 for all x and the quotient y2 /y1 has zero derivative and so is constant: y2 = c y1 for some constant c. But then y2 (x) = cy1 (x), contradicting the assump- tion that these solutions are linearly independent. Therefore k = 0 and W (x) = − for all x, as was to be shown. 7. The Wronskian of x2 and x3 is 2 W (x) = x x3 = x4 . 2x 3x2 Then W (0) = 0, while W (x) = 0 if x = 0. This is impossible if x2 and x3 are solutions of equation (2.2) for some functions p(x) and q(x). We conclude that these functions are not solutions of equation (2.2). 8. It is routine to verify that y1 (x) and y2 (x) are solutions of the differential equation. Compute W (x) = x2 = x2 . x 1 2x Then W (0) = 0 but W (x) > 0 if x = 0. However, to write the differential equation in the standard form of equation (2.2), we must divide by x2 to obtain 00 − x2 y0 + x22 y = 0. y This is undefined at x = 0, which is in the interval −1 << 1, so the theorem does not apply. 9. If y2 (x) and y2 (x) both have a relative extremum (max or min) at some x0 within (a, b), then y0 (x0 ) = y20 (x0 ) = 0. But then the Wronskian of these functions vanishes at 0, and these solu- tions must be independent. 10. By assumption, ϕ(x) is the unique solution of the initial value problem y00 + py0 + qy = 0; y(x0 ) = 0. But the function that is identically zero on I is also a solution of this initial value problem. Therefore these solutions are the same, and ϕ(x) = 0 for all x in I . 11. If y1 (x0 ) = y2 (x0 ) = 0, then the Wronskian of y1 (x) and y2 (x) is zero at x0 , and these two functions must be linearly dependent. 2.2 The Constant Coefficient Homogeneous Equa- tion 1. From the differential equation we read the characteristic equation λ2 − λ − 6 = 0, which has roots −2 and 3. The general solution is y(x) = c1 e−2x + c2 e3x . 2. The characteristic equation is λ2 − 2λ + 10 = 0 with roots 1 ± 3i. We can write a general solution y(x) = c1 ex cos(3x) + c2 ex sin(3x). 3. The characteristic equation is λ2 + 6λ + 9 = 0 with repeated roots −3, −3. Then y(x) = c1 e−3x + c2 xe−3x is a general solution. 4. The characteristic equation is λ2 − 3λ = 0 with roots 0, 3, and y(x) = c1 + c2 e3x is a general solution. 5. characteristic equation λ2 + 10λ + 26 = 0, with roots −5 ± i; general solution y(x) = c1 e−5x cos(x) + c2 e−5x sin(x). 6. characteristic equation λ2 +6λ−40 = 0, with roots 4, −10; general solution y(x) = c1 e4x + c2 e−10x . 2 +3λ+18 = 0, with roots −3/2±3 √7 i/2; general 7. characteristic equation λ solution y. 8. characteristic equation λ2 + 16λ + 64 = 0, with repeated roots −8, −8; general solution y(x) = e−8x (c1 + c2 x). 9. characteristic equation λ2 − 14λ + 49 = 0, with repeated roots 7, 7; general solution y(x) = e7x (c1 + c2 x).