Explanation Given: The differential equation is given by, y ″ ( x ) − y ′ ( x ) − 2 y ( x ) = cos x − sin 2 x ... ( 1 ) The boundary conditions are given as, y ( 0 ) = − 7 / 20 , y ′ ( 0 ) = 1 / 5 . Approach: Find the characteristic equation by substituting y ″ = r 2 y ′ = r y = 1 The complementary solution of the differential equation having real roots is, y c = c 1 e r 1 x + c 2 e r 2 x ... ( 2 ) Here, the roots of characteristic equation are r 1 , r 2 and c 1 , c 2 .are constants The general solution of the differential equation is given by, y = y c + y p ... ( 3 ) Here, y c is complementary solution and y p is particular solution of the differential equation. Calculation: The characteristic equation of equation ( 1 ) is as follows. r 2 − r − 2 = 0 Solve the above equation. r 2 − r − 2 = 0 r 2 − 2 r + r − 2 = 0 r ( r − 2 ) + 1 ( r − 2 ) = 0 ( r − 2 ) ( r + 1 ) = 0 So the roots are, r 1 = 2 r 2 = − 1 The complementary solution for the equation ( 1 ) using the roots of characteristic equation is, y c = c 1 e 2 x + c 2 e − x Let the particular solution be, y p = A cos x + B sin ( 2 x ) + C sin x + D cos ( 2 x ) ... ( 4 ) Differentiate the above equation with respect to x y ′ p = d d x ( A cos x + B sin ( 2 x ) + C sin x + D cos ( 2 x ) ) = − A sin x + 2 B cos ( 2 x ) + C cos x − 2 D sin ( 2 x ) Again differentiate the above equation with respect to x y ″ p = d d x ( − A sin x + 2 B cos ( 2 x ) + C cos x − 2 D sin ( 2 x ) ) = − A cos x − 4 B sin ( 2 x ) − C sin x − 4 D cos ( 2 x ) Substitute − A cos x − 4 B sin ( 2 x ) − C sin x − 4 D cos ( 2 x ) for y ″ , − A sin x + 2 B cos ( 2 x ) + C cos x − 2 D sin ( 2 x ) for y ′ and A cos x + B sin ( 2 x ) + C sin x + D cos ( 2 x ) for y in the equation ( 1 ) . [ − A cos x − 4 B sin ( 2 x ) − C sin x − 4 D cos ( 2 x ) − ( − A sin x + 2 B cos ( 2 x ) + C cos x − 2 D sin ( 2 x ) ) − 2 ( A cos x + B sin ( 2 x ) + C sin x + D cos ( 2 x ) ) ] = cos x − sin 2 x ( − A cos x − 4 B sin ( 2 x ) − C sin x − 4 D cos ( 2 x ) + A sin x − 2 B cos ( 2 x ) − C cos x + 2 D sin ( 2 x ) − 2 A cos x − 2 B sin ( 2 x ) − 2 C sin x − 2 D cos ( 2 x ) ) = cos x − sin ( 2 x ) ... ( 5 ) Compare the coefficients of the cos x on both sides in the above equation. − A − C − 2 A = 1 − C − 3 A = 1 C = − 3 A − 1... ( 6 ) Compare the coefficients of cos 2 x on both sides in the equation ( 5 ) . − 4 D − 2 B − 2 D = 0 2 B = − 6 D B = − 3 D ... ( 7 ) Compare the coefficients of the sin x on both sides in the equation ( 5 ) . − C + A − 2 C = 0 A = 3 C Substitute ( − 3 A − 1 ) for C in the above equation. A = 3 ( − 3 A − 1 ) A = − 9 A − 3 10 A = − 3 A = − 3 10 Substitute − 3 10 for A in equation ( 6 ) . C = − 3 ( − 3 10 ) − 1 = 9 10 − 1 = − 1 10 Compare the coefficients of the sin ( 2 x ) in the equation ( 5 )

7th Edition

ISBN: 9780321977175

Author: Nagle

Publisher: PEARSON CO

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Chapter 4.5, Problem 27E

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The solution to the initial value problem y″(x)−y′(x)−2y(x)=cosx−sin2x; y(0)=−7/20, y′(0)=1/5.

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Chapter 4.5, Problem 26E

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The solution to the initial value problem y ″ ( x ) − y ′ ( x ) − 2 y ( x ) = cos x − sin 2 x ; y ( 0 ) = − 7 / 20 , y ′ ( 0 ) = 1 / 5 .

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To find: The solution to the initial value problem y″(x)−y′(x)−2y(x)=cosx−sin2x; y(0)=−7/20, y′(0)=1/5.

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Chapter 4 Solutions

To find:

The solution to the initial value problem y″(x)−y′(x)−2y(x)=cosx−sin2x; y(0)=−7/20, y′(0)=1/5.