# ch06_4 - Ch 6.4 Differential Equations with Discontinuous...

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Ch 6.4: Differential Equations with Discontinuous Forcing Functions In this section focus on examples of nonhomogeneous initial value problems in which the forcing function is discontinuous. ( 29 ( 29 0 0 0 , 0 ), ( y y y y t g cy y b y a = = = + +

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Example 1: Initial Value Problem (1 of 12) Find the solution to the initial value problem Such an initial value problem might model the response of a damped oscillator subject to g ( t ), or current in a circuit for a unit voltage pulse. < < = - = = = = + + 20 and 5 0 , 0 20 5 , 1 ) ( ) ( ) ( where 0 ) 0 ( , 0 ) 0 ( ), ( 2 2 20 5 t t t t u t u t g y y t g y y y
Assume the conditions of Corollary 6.2.2 are met. Then or Letting Y ( s ) = L { y }, Substituting in the initial conditions, we obtain Thus )} ( { )} ( { } { 2 } { } { 2 20 5 t u L t u L y L y L y L - = + + [ ] [ ] s e e y L y y sL y sy y L s s s 20 5 2 } { 2 ) 0 ( } { ) 0 ( 2 ) 0 ( 2 } { 2 - - - = + - + - - ( 29 ( 29 ( 29 s e e y y s s Y s s s s 20 5 2 ) 0 ( 2 ) 0 ( 1 2 ) ( 2 2 - - - = - + - + + Example 1: Laplace Transform (2 of 12) ( 29 ( 29 s e e s Y s s s s 20 5 2 ) ( 2 2 - - - = + + ( 29 ( 29 2 2 ) ( 2 20 5 + + - = - - s s s e e s Y s s 0 ) 0 ( , 0 ) 0 ( ), ( ) ( 2 2 20 5 = = - = + + y y t u t u y y y

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We have where If we let h ( t ) = L -1 { H ( s )}, then by Theorem 6.3.1. ) 20 ( ) ( ) 5 ( ) ( ) ( 20 5 - - - = = t h t u t h t u t y φ Example 1: Factoring Y ( s ) (3 of 12) ( 29 ( 29 ( 29 ) ( 2 2 ) ( 20 5 2 20 5 s H e e s s s e e s Y s s s s - - - - - = + + - = ( 29 2 2 1 ) ( 2 + + = s s s s H
Thus we examine H ( s ), as follows. This partial fraction expansion yields the equations Thus Example 1: Partial Fractions (4 of 12) ( 29 2 2 2 2 1 ) ( 2 2 + + + + = + + = s s C Bs s A s s s s H 2 / 1 , 1 , 2 / 1 1 2 ) ( ) 2 ( 2 - = - = = = + + + + C B A A s C A s B A 2 2 2 / 1 2 / 1 ) ( 2 + + + - = s s s s s H

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Completing the square, Example 1: Completing the Square (5 of 12) ( 29 ( 29 ( 29 + + + + - = + + + - = + + + + - = + + + - = + + + - = 16 / 15 4 / 1 4 / 1 4 / 1 2 1 2 / 1 16 / 15 4 / 1 2 / 1 2 1 2 / 1 16 / 15 16 / 1 2 / 2 / 1 2 1 2 / 1 1 2 / 2 / 1 2 1 2 / 1 2 2 2 / 1 2 / 1 ) ( 2 2 2 2 2 s s s s s s s s s s s s s s s s s s s H
Thus and hence For h ( t ) as given above, and recalling our previous results, the solution to the initial value problem is then Example 1: Solution (6 of 12) ( 29 ( 29 ( 29 ( 29 ( 29 + + - + + + - = + + + + - = 16 / 15 4 / 1 4 / 15 15 2 1 16 / 15 4 / 1 4 / 1 2 1 2 / 1 16 / 15 4 / 1 4 / 1 4 / 1 2 1 2 / 1 ) ( 2 2 2 s s s s s s s s H - - = = - - - t e t e s H L t h t t 4 15

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