Explanation Procedure used: 1. Make the characteristic equation by substituting y ″ = r 2 , y ′ = r . The characteristic equation is, a r 2 + b r + c = 0 2. Find the roots of the characteristic equation. 3. If roots are real. Case-1 If r = r 1 = r 2 , then the solution is y h ( t ) = c 1 e r t + c 2 t e r t Case-2 If r 1 ≠ r 2 , then the solution is y h ( t ) = c 1 e r 1 t + c 2 e r 2 t 4. If roots are imaginary in the form of r 1 , 2 = α ± β i , then the solution is y h ( t ) = e a t ( c 1 cos β t + c 2 sin β t ) For each function f i ( t ) in f ( t ) , determine the associated set of functions F i and F i can be obtained as, 5. If f i ( t ) = t m then the associated function is, F = { 1 , t , t 2 , ... , t m } 6. If f i ( t ) = t m e α t then the associated function is, F = { e α t , e α t t , e α t t 2 , ... , e α t t m } 7. If f i ( t ) = t m e α t cos β t or f i ( t ) = t m e α t sin β t then the associated function 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 } The particular solution can be obtained by taking the linear combination of all functions in the associated sets. If any repeated function present then it should appear only once in the particular solution. The general solution is given by, y ( t ) = y h ( t ) + y p ( t ) 8. Substitute boundary conditions to get the values of constant. 9. Substitute values of constant in the general solution to get the final solution. Calculation: Consider the given differential equation. y ″ + 9 y = sin 3 t (1) The characteristic equation is, r 2 + 9 = 0 The roots of the above equation are, r 2 + 9 = 0 r 2 = − 9 r = ± 3 i Here, the roots are imaginary. Substitute 0 for α and 3 for β in general solution. y h ( t ) = c 1 cos 3 t + c 2 sin 3 t Consider the forcing function. f ( t ) = sin 3 t The form of the function is r 2 + 9 with m = 0 . The associated set is F 1 = { sin 3 t , cos 3 t } . Since the function is a solution to the corresponding homogeneous equation. Hence, the particular solution is a linear combination of function F , the solution is given as, y p = A t sin 3 t + B t cos 3 t (2) Differentiate Equation (2) With respect to t

5th Edition

ISBN: 9780128149485

Author: Abell, Martha L. L.

Publisher: Elsevier Science

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Differential Equations

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Chapter 4, Problem 44RE

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The solution of the initial value problem y″+9y=sin3t with initial condition y(0)=6, y′(0)=0

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Chapter 4, Problem 43RE

Chapter 4, Problem 45RE

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Chapter 4 Solutions

Introductory Differential Equations

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The solution of the initial value problem y ″ + 9 y = sin 3 t with initial condition y ( 0 ) = 6 , y ′ ( 0 ) = 0

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The solution of the initial value problem y″+9y=sin3t with initial condition y(0)=6, y′(0)=0

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Chapter 4 Solutions