Explanation Procedure used: 1. Make 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 general 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 general solution is y h ( t ) = c 1 e r 1 t + c 2 e r 2 t . 4. If roots are imaginary in form of r 1 , 2 = α ± β i , then the general solution is y h ( t ) = e a t ( c 1 cos β t + c 2 sin β t ) . 5. Find particular solution y p ( t ) . 6. If find the fundamental set S = { y 1 ( t ) , y 2 ( t ) } is, then the wronskian is, W ( s ) = | y 1 ( t ) y 2 ( t ) d d t ( y 1 ( t ) ) d d t ( y 2 ( t ) ) | 7. Calculate u 1 ′ and u 2 ′ as, u ′ 1 = − y 2 f ( t ) W ( s ) and u ′ 2 = y 1 f ( t ) W ( s ) 8. Integrate u 1 ′ and u 2 ′ to get the values of u 1 and u 2 . 9. Particular solution is given as y p ( t ) = u 1 y 1 + u 2 y 2 . 10. The solution of differential equation is given as y ( t ) = y h ( t ) + y p ( t ) . Calculation: Consider the differential equation. y ″ + 9 y = sec 3 t tan 3 t The characteristic equation is, r 2 + 9 = 0 . The roots of the above equation is, r 2 + 9 = 0 r 2 = − 9 r = ± 3 i Here, roots are imaginary. Substitute 0 for α and 3 for β in general solution for imaginary roots. y h ( t ) = e 0 × t ( c 1 cos ( 3 t ) + c 2 sin ( 3 t ) ) y h ( t ) = c 1 cos 3 t + c 2 sin 3 t The forcing function is f ( t ) = sec 3 t tan 3 t . The fundamental solution is S = { cos 3 t , sin 3 t } . Here, y 1 ( t ) = cos 3 t and y 2 = sin 3 t . The wronskian is, W ( s ) = | cos 3 t sin 3 t d d t ( cos 3 t ) d d t ( sin 3 t ) | = | cos 3 t sin 3 t − 3 sin 3 t 3 cos 3 t | = ( 3 cos 3 t ) cos 3 t − sin 3 t ( − 3 sin 3 t ) = 3 Substitute values of y 2 ( t ) , f ( t ) and W ( S ) to calculate u ′ 1

5th Edition

ISBN: 9780128149485

Author: Abell, Martha L. L.

Publisher: Elsevier Science

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Differential Equations

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Chapter 4.4, Problem 38E

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The solution of y″+9y=sec3ttan3t,−π6<t<π6.

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Chapter 4.4, Problem 37E

Chapter 4.4, Problem 39E

Chapter 4 Solutions

Introductory Differential Equations

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The solution of y ″ + 9 y = sec 3 t tan 3 t , − π 6 &lt; t &lt; π 6 .

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The solution of y″+9y=sec3ttan3t,−π6<t<π6.

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Chapter 4 Solutions

The solution of y ″ + 9 y = sec 3 t tan 3 t , − π 6 < t < π 6 .