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Transcript
Solutions 2.3-Page 131 Problem 1 Find the general solutions of the differential equations. y ′′ − 4 y = 0 The first step is to find the roots of the characteristic equation. r2 − 4 = 0 (r + 2)(r − 2) = 0 r = −2,2 Since the roots are real and distinct, the general solution is y ( x) = c1e −2 x + c 2 e 2 x Problem 9 Find the general solutions of the differential equations. y ′′ + 8 y ′ + 25 y = 0 The characteristic equation is r 2 + 8r + 25 = 0 . The roots can be found using the quadratic equation formula or a calculator. The roots are r = −4 ± 3i . The general solution is based on Theorem 3 on page 128. y ( x) = e −4 x (c1 cos 3x + c 2 sin 3 x) Problem 21 Solve the initial value problems. y′′ − 4 y′ + 3 y = 0; y (0) = 7, y′(0) = 11 The characteristic equation is r 2 − 4r + 3 = 0 . The roots can be found using the quadratic equation formula or a calculator. The roots are r = 3,1 . The general solution is y = c1e 3 x + c 2 e x . The initial conditions are used to find the constants. y ′ = 3c1e 3 x + c 2 e x . Substituting the initial conditions gives: y (0) = 7 = c1 + c 2 y ′(0) = 11 = 3c1 + c 2 Solving the system of equations gives c1 = 2, c 2 = 5 y ( x) = 2e 3 x + 5e x Problem 32 Find the general solutions of the equations. First find a small integral root of the characteristic equation by inspection; then factor by division. y ( 4 ) + y ( 3) − 3 y ′′ − 5 y ′ − 2 y = 0 The characteristic equation is r 4 + r 3 − 3r 2 − 5r − 2 = 0 . The root obtained from inspection is r = 2. Long division of the characteristic equation by (r-2) yields: (r + 1) 3 = 0 With the root r=-1 repeated three times, the general solution is y ( x) = c1e 2 x + (c 2 + c3 x + c 4 x 2 )e − x