Explanation Given: The differential equation is, θ ″ ( t ) − θ ( t ) = t sin t Approach: 1. Find the characteristic equation by substituting y ″ = r 2 , y ′ = r . 2. Find the roots of the characteristic equation. 3. If roots are real. Case-1 The general solution if r = r 1 = r 2 is, y ( t ) = c 1 e r t + c 2 t e r t Case-2 The general solution if r 1 ≠ r 2 is, y ( t ) = c 1 e r 1 t + c 2 e r 2 t 4. The general solution if roots are imaginary in form of r 1 , 2 = α ± β i is, y ( t ) = e a t ( c 1 cos β t + c 2 sin β t ) For each function in f i ( t ) in f ( t ) , determine the associated set of functions F i . 5. F i can be obtained as, The associate function if f i ( t ) = t m is, F = { 1 , t , t 2 , ... , t m } The associate function if f i ( t ) = t m e α t is, F = { e α t , e α t t , e α t t 2 , ... , e α t t m } The associate function if f i ( t ) = t m e α t cos β t or f i ( t ) = t m e α t sin β t is, F = { e α t cos β t , e α t t cos β t , e α t t 2 cos β t , ... , e α t t m cos β t , e α t sin β t , e α t t sin β t , e α t t 2 sin β t , ... , e α t t m sin β t } 6. The general solution if roots are imaginary in form of r 1 , 2 = α ± β i is, y ( t ) = e a t ( c 1 cos β t + c 2 sin β t ) The particular solution can be obtained by taking the linear combination of all functions in the associated sets. If any repeated function is present then it should appear only once in the particular solution. Calculation: Consider the given equation. θ ″ ( t ) − θ ( t ) = t sin t … ( 1 ) The auxiliary equation is written as, r 2 − 1 = 0 r = ± 1 r = − 1 , 1 As f ( t ) = t sin t its associated roots will be complex. The complementary solution is, y ( t ) = e a t ( c 1 cos ( 1 ) t + c 2 sin ( − 1 ) t ) Now, calculate the particular solution, Take θ p = ( A t + B ) sin t + ( C t + D ) cos t … ( 2 ) Differentiate the above equation with respect to t , θ ′ p = ( A t + B ) cos t + sin t ( A ) + ( C t + D ) ( − sin t ) + cos t ( C ) θ ″ p = ( A t + B ) ( − sin t ) + cos t ( A ) + A cos t + ( C t + D ) ( − cos t ) + ( − sin t ) ( C ) − C sin t Substitute the value of θ ′ p and θ ″ p in the equation ( 1 ) θ ″ ( t ) − θ ( t ) = t sin t

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 4.4, Problem 16E

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The particular solution of the differential equation θ″(t)−θ(t)=tsint.

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Chapter 4.4, Problem 15E

Chapter 4.4, Problem 17E

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The particular solution of the differential equation θ ″ ( t ) − θ ( t ) = t sin t .

Solution Summary: The author explains how to find the particular solution of the differential equation theta

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The particular solution of the differential equation θ″(t)−θ(t)=tsint.

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The particular solution of the differential equation θ ″ ( t ) − θ ( t ) = t sin t .

Solution Summary: The author explains how to find the particular solution of the differential equation theta

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To find: The particular solution of the differential equation θ″(t)−θ(t)=tsint.

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Chapter 4 Solutions

To find:

The particular solution of the differential equation θ″(t)−θ(t)=tsint.