To determine To solve: The boundary value problem y ″ + y = 0 , y ( 0 ) = 0 , y ′ ( π ) = 1 . Explanation The given differential equation is y ″ + y = 0 . The characteristic equation corresponding to y ″ + y = 0 is m 2 + 1 = 0 . Compute the roots of m 2 + 1 = 0 as shown below. m 2 + 1 = 0 m 2 = − 1 m = ± i The roots are m = + i and − i . Since the characteristic equation has complex conjugate roots, the general solution of y ″ + y = 0 is given by y = c 1 cos x + c 2 sin x . Now use the given initial conditions y ( 0 ) = 0 , y ′ ( π ) = 1 and evaluate c 1 and c 2 . Here, the complementary function is y ( x ) = c 1 cos x + c 2 sin x . Substitute x = 0 and y ( 0 ) = 0 in the complementary function y ( x ) = c 1 cos x + c 2 sin x as follows. y ( x ) = c 1 cos x + c 2 sin x y ( 0 ) = c 1 cos 0 + c 2 sin 0 0 = c 1 ⋅ 1 + c 2 ⋅ 0 0 = c 1 Therefore, the value of c 1 is 0. Compute the derivative of the function y ( x ) = c 1 cos x + c 2 sin x as shown below

Elementary Differential Equations and Boundary Value Problems, Enhanced

11th Edition

ISBN: 9781119381648

Author: Boyce

Publisher: WILEY

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Domain

In simple mathematics terms, a domain is defined as the space on which all the possible inputs for the function will lie. It can also be expressed as a set of departures for the required function.

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Chapter 10.1, Problem 1P

To determine

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

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Chapter 9.8, Problem 15P

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Elementary Differential Equations and Boundary Value Problems, Enhanced

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