Advanced Engineering Mathematics 8th Edition ONeil SOLUTIONS MANUAL AND TESTBANK

Chapter 2 Second-Order Differential Equations 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).