nagle_differential_equations_ISM_Part48

# nagle_differential_equations_ISM_Part48 - Exercises 7.6 Let...

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Unformatted text preview: Exercises 7.6 Let Y ( s ) = L{ y } ( s ). Applying the Laplace transform to the given equation and using the initial conditions, we obtain L{ y 00 } ( s ) + 2 L{ y } ( s ) + 10 Y ( s ) = L{ g ( t ) } ( s ) ⇒ s 2 Y ( s ) + s + 2 [ sY ( s ) + 1] + 10 Y ( s ) = 10 s ( 1 + e- 10 s- 2 e- 20 s ) ⇒ Y ( s ) =- s + 2 ( s + 1) 2 + 9 + 10 s [( s + 1) 2 + 9] ( 1 + e- 10 s- 2 e- 20 s ) . Using partial fractions decomposition, we can write Y ( s ) =- s + 2 ( s + 1) 2 + 9 + 1 s- s + 2 ( s + 1) 2 + 9 ( 1 + e- 10 s- 2 e- 20 s ) = 1 s- 2( s + 2) ( s + 1) 2 + 9 + 1 s- s + 2 ( s + 1) 2 + 9 ( e- 10 s- 2 e- 20 s ) = 1 s- 2 s + 1 ( s + 1) 2 + 9- 2 3 3 ( s + 1) 2 + 9 + 1 s- s + 1 ( s + 1) 2 + 9- 1 3 3 ( s + 1) 2 + 9 ( e- 10 s- 2 e- 20 s ) . Therefore, taking the inverse Laplace transform, we finally obtain y ( t ) = 1- 2 cos 3 te- t- 2 3 sin 3 te- t + 1- e- ( t- 10) cos 3( t- 10)- 1 3 e- ( t- 10) sin 3( t- 10) u ( t- 10)- 2 1- e- ( t- 20) cos 3( t- 20)- 1 3 e- ( t- 20) sin 3( t- 20) u ( t- 20) . 40. We can express g ( t ) using the unit step function as g ( t ) = e- t + ( 1- e- t ) u ( t- 3) . Thus, taking the Laplace transform yields L{ g ( t ) } ( s ) = 1 s + 1 + 1 s- e- 3 s + 1 e- 3 s so that L{ y 00 + 3 y + 2 y } ( s ) = s 2 Y ( s )- 2 s + 1 + 3 [ sY ( s )- 2] + 2 Y ( s ) = 1 s + 1 + 1 s- e- 3 s + 1 e- 3 s , where Y ( s ) = L{ y } ( s ). Solving for Y ( s ), we obtain ( s 2 + 3 s + 2 ) Y ( s ) = 2 s + 5 + 1 s + 1 + 1 s- e- 3...
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nagle_differential_equations_ISM_Part48 - Exercises 7.6 Let...

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