Explanation The given differential equation is 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 general solution of y ″ + 2 y = 0 is given by y = c 1 cos 2 x + c 2 sin 2 x . Now use the given initial conditions y ′ ( 0 ) = 1 , y ′ ( π ) = 0 and evaluate c 1 and c 2 . Here, the complementary function is y ( x ) = c 1 cos 2 x + c 2 sin 2 x . Compute the derivative of the function y ( x ) = c 1 cos 2 x + c 2 sin 2 x as shown below. y ( x ) = c 1 cos 2 x + c 2 sin 2 x y ′ ( x ) = c 1 ( − 2 sin 2 x ) + 2 c 2 cos 2 x y ′ ( x ) = − 2 c 1 sin 2 x + 2 c 2 cos 2 x Substitute x = 0 and y ′ ( 0 ) = 1 in the differential equation y ′ ( x ) = − 2 c 1 sin 2 x + 2 c 2 cos 2 x as follows. y ′ ( x ) = − 2 c 1 sin 2 x + 2 c 2 cos 2 x y ′ ( 0 ) = − 2 c 1 sin ( 2 ⋅ 0 ) + 2 c 2 cos ( 2 ⋅ 0 ) 1 = − 2 c 1 sin 0 + 2 c 2 cos 0 1 = − 2 c 1 ( 0 ) + 2 c 2 ( 1 ) Simplify further to get c 2 as follows. 2 c 2 = 1 c 2 = 1 2 c 2 = 2 2 Therefore, the value of c 2 is 2 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 2P

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To solve: The boundary value problem y″+2y=0, y′(0)=1, y′(π)=0.

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Elementary Differential Equations and Boundary Value Problems, Enhanced

Ch. 10.1 - In each of Problems 1 through 13, either solve the...Ch. 10.1 - Prob. 2P Ch. 10.1 - In each of Problems 1 through 13, either solve the...Ch. 10.1 - Prob. 4P Ch. 10.1 - In each of Problems 1 through 13, either solve the...Ch. 10.1 - In each of Problems 1 through 13, either solve the...Ch. 10.1 - In each of Problems 1 through 13, either solve the...Ch. 10.1 - Prob. 8P Ch. 10.1 - Prob. 9P Ch. 10.1 - In each of Problems 1 through 13, either solve the...

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The boundary value problem y ″ + 2 y = 0 , y ′ ( 0 ) = 1 , y ′ ( π ) = 0 .

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To solve: The boundary value problem y″+2y=0, y′(0)=1, y′(π)=0.

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