Explanation Procedure used: Particular solution to a linear DE by greens function. 1. Consider the differential equation and determine the complementary function to the homogeneous differential equation. Further, substitute the boundary values to find the coefficients. 2. If y 1 ( x ) and y 2 ( x ) are two fundamental solution of the differential equation then particular solution is given by y p = u 1 y 1 ( x ) + u 2 y 2 ( x ) . 3. Use, y p = ∫ x 0 x G ( x , t ) f ( t ) d t to find the particular solution. Find greens function from G ( x , t ) = y 1 ( t ) y 2 ( x ) W t . 4. Obtain the final solution by y ( x ) = y h + y p . Calculation: Consider the differential equation y ″ + 9 y = 1 . The associated homogeneous differential equation is y ″ + 9 y = 0 . The auxiliary equation to the homogeneous DE y ″ + 9 y = 0 is m 2 + 9 = 0 . Find the roots of m 2 + 9 = 0 . m 2 + 9 = 0 m 2 = − 9 Thus, the roots of the auxiliary equation are m 1 = 3 i and m 2 = − 3 i Hence, the solution of homogeneous DE y ″ + 9 y = 0 is y h = c 1 cos 3 x + c 2 sin 3 x . Consider the boundary values y ( 0 ) = 0 , y ( π ) = 0 . Substitute y ( 0 ) = 0 into y h = c 1 cos 3 x + c 2 sin 3 x . Thus, 0 = c 1 Thus, choose c 2 = 1 . Hence, y 1 = sin 3 x . Substitute y ( π ) = 0 . Since y ′ h = − 3 c 1 sin 3 x + 3 c 2 cos 3 x . − 3 c 2 = 0 c 2 = 0 Choose, c 1 = 1 . Thus, y 2 = cos 3 x . The particular solution to the nonhomogeneous differential equation is given by y p = ∫ x 0 x G ( x , t ) f ( t ) d t . Now, determine the Wronskian. Use, variable transformation. Thus, [ y 1 ( x ) = sin 3 x defines y 1 ( t ) = sin 3 t , y 2 ( x ) = cos 3 x defines y 2 ( t ) = cos 3 t f ( x ) = 1 defines f ( t ) = 1 ] W ( t ) = | y 1 ( t ) y 2 ( t ) y ′ 1 ( t ) y ′ 2 ( t ) | = | sin 3 t cos 3 t 3 cos 3 t − 3 sin 3 t | = − 3 ( sin 2 3 t + cos 2 3 t ) [ Since sin 2 θ + cos 2 θ = 1 ] = − 3 Thus, for any t W ( t ) ≠ 0

Differential Equations with Boundary-Value Problems (MindTap Course List)

9th Edition

ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 4.8, Problem 40E

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The solution of the given boundary values problem $y″+9y=1,y(0)=0,y(π)=0$ .

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Chapter 4.8, Problem 39E

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Differential Equations with Boundary-Value Problems (MindTap Course List)

Ch. 4.1 - In Problems 14 the given family of functions is...Ch. 4.1 - In Problems 14 the given family of functions is...Ch. 4.1 - In Problems 14 the given family of functions is...Ch. 4.1 - In Problems 14 the given family of functions is...Ch. 4.1 - Given that y=c1+c2x2 is a two-parameter family of...Ch. 4.1 - Find two members of the family of solutions in...Ch. 4.1 - Prob. 7E Ch. 4.1 - Use the general solution of x + 2x = 0 given in...Ch. 4.1 - In Problems 9 and 10 find an interval centered...Ch. 4.1 - In Problems 9 and 10 find an interval centered...

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The solution of the given boundary values problem y ″ + 9 y = 1 , y ( 0 ) = 0 , y ( π ) = 0 .

Solution Summary: The author explains the solution of the boundary value problem y′′+9y=1 and the auxiliary equation to the homogeneous DE.

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The solution of the given boundary values problem $y″+9y=1,y(0)=0,y(π)=0$ .

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