Explanation Given: The boundary value problem is, y ″ + λ 2 y = sin t ... ( 1 ) The boundary conditions are given as y ( 0 ) = 0 and y ( π ) = 1 . Approach: The complementary solution for the differential equation if the roots of the auxiliary function are imaginary is given by, y c = c 1 e α t cos β t + c 2 e α t sin β t Here, the roots of the auxiliary equation are α + i β and α − i β and c 1 , c 2 are arbitrary constants. The general solution of the differential equation is given by, y = y c + y p Here, y p is the particular solution and y c is the complementary solution. Calculation: The auxiliary equation of equation ( 1 ) is as follows. r 2 + λ 2 = 0 Solve the above equation using the solution of quadratic equation. r 2 + λ 2 = 0 r 2 = − λ 2 r = ± λ i So the roots are, r 1 = λ i r 2 = − λ i The complementary solution is as follows. y c = c 1 cos λ t + c 2 sin λ t Assume the particular solution of the differential equation as follows. y p = A sin t + B cos t ... ( 3 ) Differentiate the above equation with respect to t . y ′ = d d t ( A sin t + B cos t ) = A cos t − B sin t Again differentiate the above equation with respect to t . y ″ = d d t ( A cos t − B sin t ) = − A sin t − B cos t Substitute − A sin t − B cos t for y ″ and A sin t + B cos t for y in the equation ( 1 ) . − A sin t − B cos t + λ 2 ( A sin t + B cos t ) = sin t − A sin t − B cos t + λ 2 A sin t + λ 2 B cos t = sin t ... ( 3 ) Compare the coefficients of sin t on both sides in the above equation. λ 2 A − A = 1 A = 1 λ 2 − 1 Compare the coefficients of the cos t on both sides in the equation ( 3 )

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 4.5, Problem 46E

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The boundary value problem has solution if and only if λ≠±1,±2,±3,....

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Chapter 4.5, Problem 45E

Chapter 4.5, Problem 47E

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The boundary value problem has solution if and only if λ ≠ ± 1 , ± 2 , ± 3 , ... .

Solution Summary: The author explains that the boundary value problem has solution only if the roots of the auxiliary function are imaginary.

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The boundary value problem has solution if and only if λ≠±1,±2,±3,....

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The boundary value problem has solution if and only if λ ≠ ± 1 , ± 2 , ± 3 , ... .

Solution Summary: The author explains that the boundary value problem has solution only if the roots of the auxiliary function are imaginary.

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To show: The boundary value problem has solution if and only if λ≠±1,±2,±3,....

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Chapter 4 Solutions

To show:

The boundary value problem has solution if and only if λ≠±1,±2,±3,....