To determine The solution of the initial value problem y ″ − 3 y ′ + 2 y = 10 sin x , y ( 0 ) = 1 , y ′ ( 0 ) = − 6 and graph it. Explanation Result used: The general solution of Euler-Cauchy equation: “Suppose that, the linear ordinary differential equation of the form x 2 y ″ + a x y ′ + b y = 0 . In order to obtain the auxiliary equation, substitute y = x m , y ′ = m x m − 1 and y ″ = m ( m − 1 ) x m − 2 . The auxiliary equation is m 2 + ( a − 1 ) m + b = 0 . Then, then the general solution of the equation is of the form: Case 1: If the roots are real distinct m 1 and m 2 , then the general solution is, y h ( x ) = c 1 x m 1 + c 2 x m 2 . Case 2: If the roots are real double roots m 1 and m 1 , then the general solution is, y h ( x ) = ( c 1 + c 2 ln x ) x m 1 . Case 3: If the roots are complex a ± i b , then the general solution is, y h ( x ) = x a ( A cos ( b ln x ) + B sin ( b ln x ) ) . Calculation: Given that, the ordinary differential equation is y ″ − 3 y ′ + 2 y = 10 sin x . First compute the general solution y h of the nonhomogeneous linear ordinary differential equation as follows. The auxiliary equation of y ″ − 3 y ′ + 2 y = 10 sin x is m 2 − 3 m + 2 = 0 . m 2 − 3 m + 2 = 0 m 2 − m − 2 m + 2 = 0 ( m − 1 ) ( m − 2 ) = 0 m = 1 or 2 Thus, the roots of the auxiliary equation m 2 − 3 m + 2 = 0 are 1 and 2 . Here, the roots are real distinct. So, the general solution y h of the nonhomogeneous linear ordinary differential equation is, y h ( x ) = A e x + B e 2 x . Now compute the particular solution of differential equation y ″ − 3 y ′ + 2 y = 10 sin x as follows. Since the right hand side function is of the form k sin w x , the choice for particular solution is K cos w x + M sin w x . Therefore, the particular solution of y ″ − 3 y ′ + 2 y = 10 sin x becomes y p = K cos x + M sin x . Differentiate y p = K cos x + M sin x with respect to x as shown below. y ′ p = d d x ( K cos x + M sin x ) = K ( − sin x ) + M cos x = − K sin x + M cos x Again differentiate y ′ p with respect to x and obtain y ″ p as follows. y ″ p = d d x ( − K sin x + M cos x ) = − K ( cos x ) + M ( − sin x ) = − K cos x − M sin x Since the particular solution satisfies the given ordinary differential equation, substitute y p , y ′ p , and y ″ p in y ″ − 3 y ′ + 2 y = 10 sin x as shown below. y ″ − 3 y ′ + 2 y = 10 sin x { ( − K cos x − M sin x ) − 3 ( − K sin x + M cos x ) + 2 ( K cos x + M sin x ) } = 10 sin x { − K cos x − M sin x + 3 K sin x − 3 M cos x + 2 K cos x + 2 M sin x } = 10 sin x { ( K − 3 M ) cos x + ( 3 K + M ) sin x } = 10 sin x Equate the coefficients of like terms on both sides as follows. K − 3 M = 0 … … ( 1 ) 3 K + M = 10 … … ( 2 ) From equation (1), the value of K = 3 M

10th Edition

ISBN: 9781119455929

Author: Kreyszig

Publisher: WILEY

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Chapter 2, Problem 20RQ

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The solution of the initial value problem y″−3y′+2y=10sinx,y(0)=1,y′(0)=−6 and graph it.

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