To determine To solve: The boundary value problem y ″ + 2 y = x , y ( 0 ) = 0 , y ( π ) = 0 . Explanation The given differential equation is y ″ + 2 y = x . Consider the corresponding homogeneous differential equation y ″ + 2 y = 0 . The characteristic equation corresponding to y ″ + 2 y = 0 is m 2 + 2 = 0 . Compute the roots of m 2 + 2 = 0 as shown below. m 2 + 2 = 0 m 2 = − 2 m = ± 2 i The roots are m = + 2 i and − 2 i . Since the characteristic equation has complex conjugate roots, the complementary function is given by y h = c 1 cos 2 x + c 2 sin 2 x . Compute the particular integral of the nonhomogeneous differential equation y ″ + 2 y = x as shown below. The undetermined coefficient set corresponding to the right hand side function x is { 1 , x } . Consider the linear combination y = d 1 + d 2 x of 1 and x . Compute the first and the second derivatives of the equation y = d 1 + d 2 x . y = d 1 + d 2 x y ′ = d 2 y ″ = 0 Substitute for y , y ′ and y ″ in the differential equation y ″ + 2 y = x . y ″ + 2 y = x 0 + 2 ( d 1 + d 2 x ) = x 2 ( d 1 + d 2 x ) = x 2 d 1 + 2 d 2 x = x Compare the coefficients of x and constant terms on both sides of the above equation. d 2 = 1 2 , d 1 = 0 Substitute d 1 = 0 and d 2 = 1 2 in y = d 1 + d 2 x . Therefore, y p = x 2 is the particular integral of the differential equation y ″ + 2 y = x . The general solution of y ″ + 2 y = x is given by y = y h + y p . Hence, y = c 1 cos 2 x + c 2 sin 2 x + x 2 is the general solution of the differential equation y ″ + 2 y = x . Use the given initial conditions y ( 0 ) = 0 , y ( π ) = 0 to evaluate c 1 and c 2

Elementary Differential Equations and Boundary Value Problems, Enhanced

11th Edition

ISBN: 9781119381648

Author: Boyce

Publisher: WILEY

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Chapter 10.1, Problem 6P

To determine

To solve: The boundary value problem y″+2y=x, y(0)=0, y(π)=0.

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Chapter 10.1, Problem 5P

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Elementary Differential Equations and Boundary Value Problems, Enhanced

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