set3_solutions

set3_solutions - Sample for the Midterm 2 for MA527 1. Find...

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Unformatted text preview: Sample for the Midterm 2 for MA527 1. Find the inverse Laplace transform of s 2 +6 ( s- 2)( s 2 +2 s +2) . Solution : Decompose the rational function into two: s 2 + 6 ( s- 2)( s 2 + 2 s + 2) = A s- 2 + Bs + C s 2 + 2 s + 2 . A, B, C are easily computable by clearing fractions and comparing the numerators: s 2 + 6 = A ( s 2 + 2 s + 2) + ( Bs + C )( s- 2) First, substitute s = 2. Then we get 10 = 10 A . Comparing the coefficient of s 2 , we get B = 0, and comparing the constants, we get C =- 2. Since s 2 + 2 s + 2 = ( s + 1) 2 + 1, we have an s-shift, and hence the answer is L- 1 ( 1 s- 2 )- 2 L- 1 ( 1 s 2 + 2 s + 2 ) = e 2 t- 2 e- t sin t 2. Find the inverse Laplace transform of e- 3 s s 2 +4 s +5 . Solution : First, L- 1 ( 1 s 2 + 4 s + 5 ) = L- 1 ( 1 ( s + 2) 2 + 1 ) = e- 2 t sin t. Therefore by the second shift theorem the answer is equal to e- 2( t- 3) sin( t- 3) u ( t- 3) 3. Compute u ( t- 1) * ( e- 2 t u ( t )) and its Laplace transform. Solution : By definition, u ( t- 1) * ( e- 2 t u ( t )) = Z t u ( τ- 1) e- 2( t- τ ) u ( t- τ ) dτ Note that if τ < 1 then the first factor under the integral sign is zero. Hence if t < 1, then τ < 1, so the integral is 0. If t ≥ 1, then still the integrand is 0 for τ < 1 and reduces to e- 2( t- τ ) for 1 < τ < t . Hence in this case the integral is equal to e- 2 t R t 1 e 2 τ dτ = 1 2 (1- e 2- 2 t ). Therefore, the final answer is u ( t- 1) * ( e- 2 t u ( t )) = 1 2 (1- e 2- 2 t ) u ( t- 1) The Laplace transform is equal to the product of Laplace transforms, i.e. to L ( u ( t- 1)) L ( e- 2 t ) = e- s s ( s +2) . 4. Solve y ( t ) = 2 t- 4 R t y ( τ )( t- τ )d τ . Solution : This means y = 2 t- 4 y * t and taking Laplace transform we get Y = 2 s 2- 4 Y · 1 s 2 . Therefore Y = 2 s 2 +4 and y = L- 1 ( 2 s 2 +4 ) = sin 2 t . 5. Solve y 00 + 2 y- 3 y = 8 e- t + δ ( t- 1 / 2) y (0) = 3 y (0) =- 5. Solution : Taking Laplace transform, one gets ( s 2 + 2 s- 3) Y = 8 s + 1 + e- s 2 + ( s + 2) y (0) + y (0) i.e. Y = 8 ( s + 1)( s- 1)( s + 3) + e- s/ 2 ( s- 1)( s + 3) + 3 s + 1 ( s- 1)( s + 3) ....
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This note was uploaded on 04/03/2012 for the course MA 527 taught by Professor Weitsman during the Spring '08 term at Purdue University.

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set3_solutions - Sample for the Midterm 2 for MA527 1. Find...

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