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 r is real and k ≥ 2 , then The solution associated with r and k is e r t , t e r t , t 2 e r t , ... , t k − 1 e r t . 4. If r = α ± β i then the solution associated are e α t cos β t and e α t sin β t . 5. Find the particular solution y p ( t ) . 6. find the fundamental set as: If S = { y 1 ( t ) , y 2 ( t ) } Then, the wronskian is W ( s ) = | y 1 ( t ) y 2 ( t ) y 3 ( t ) d d t ( y 1 ( t ) ) d d t ( y 2 ( t ) ) d d t ( y 3 ( t ) ) d 2 d t 2 ( y 1 ( t ) ) d 2 d t 2 ( y 2 ( t ) ) d 2 d t 2 ( y 3 ( t ) ) | . 7. Calculate u 1 ′ u 2 ′ and u 3 ′ as, u k ′ = { a i j | a i j ∈ W ( s ) except a i k = − c } 8. Integrate u 1 ′ u 2 ′ and u 3 ′ to get the values of u 1 , u 2 and u 3 . 9. The particular solution is given as y p ( t ) = u 1 y 1 + u 2 y 2 + u 3 y 3 . Here, y 1 , y 2 , and y 2 are fundamental solutions of the differential equation. 10. The solution of a differential equation is given as y ( t ) = y h ( t ) + y p ( t ) . Calculation: Consider the differential equation y ‴ + y ′ = − sec t tan t (1) The characteristic equation is r 3 + r = 0 . The roots of the above equation are, r 3 + r = 0 r ( r 2 + 1 ) = 0 Here, r = 0 , k = 1 , and r = ± i , k = 1 For real root the fundamental set with respect to r = 0 , k = 1 , S 1 = e 0 × t S 1 = 1 For imaginary root the fundamental set with respect to r = ± 2 i , k = 1 , S 2 = e 0 × t cos t , e 0 × t sin t S 2 = cos t , sin t The fundamental set of solution is S = { 1 , cos t , sin t } . Here, y 1 ( t ) = 1 , y 2 ( t ) = cos t and y 2 = sin t The general solution is y h ( t ) = c 1 + c 2 cos t + c 3 sin t (2) The Wronskian is, W ( s ) = | 1 cos t sin t d d t ( 1 ) d d t ( cos t ) d d t ( sin t ) d 2 d t 2 ( 1 ) d 2 d t 2 ( cos t ) d 2 d t 2 ( sin t ) | = | 1 cos t sin t 0 − sin t cos t 0 − cos t − sin t | = 1 ( sin 2 t + cos 2 t ) = 1 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 4.6, Problem 16E

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To find: The solution of y‴+y′=−secttant,−π2<t<π2 by method of parameters.

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Chapter 4.6, Problem 13E

Chapter 4.6, Problem 17E

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Chapter 4 Solutions

Introductory Differential Equations

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The solution of y ‴ + y ′ = − sec t tan t , − π 2 &lt; t &lt; π 2 by method of parameters.

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To find: The solution of y‴+y′=−secttant,−π2<t<π2 by method of parameters.

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Chapter 4 Solutions

The solution of y ‴ + y ′ = − sec t tan t , − π 2 < t < π 2 by method of parameters.