202s04mt2

# 202s04mt2 - B U Department of Mathematics Math 202...

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Unformatted text preview: B U Department of Mathematics Math 202 Differential Equations Date: May 14, 2004 Full Name : Time: 18:10-19:30 Math 202 Number : Student ID : Spring 2004 Second Midterm – Solution Key IMPORTANT 1. Write your name, surname on top of each page. 2. The exam consists of 4 questions some of which have more than one part. 3. Read the questions carefully and write your answers neatly under the corresponding questions. 4. Show all your work. Correct answers without sufficient explanation might not get full credit. 5. Calculators are not allowed. Q1 Q2 Q3 Q4 total 25 pts 25 pts 30 pts 25 pts 105 pts Some basic Laplace transforms you might need L { 1 } = 1 s , s > L { e at } = 1 s- a , s > a L { sin at } = a s 2 + a 2 , s > L { cos at } = s s 2 + a 2 , s > L { sinh at } = a s 2- a 2 , s > | a | L { cosh at } = s s 2- a 2 , s > | a | 1.) [25] Use the Laplace Transform to solve the initial value problem: y 00- 2 y + 5 y = g ( t ) where g ( t ) = ≤ t < 1 e t 1 ≤ t subject to the initial conditions y (0) = y (0) = 0. Is the solution continuous at t = 1? Solution: Using the step function representation: y 00- 2 y + 5 y = u 1 ( t ) e t . Taking the Laplace Transform of both sides we get: L { y 00- 2 y + 5 y } = L { u 1 ( t ) e t } ⇒ s 2 Y ( s )- sy (0)- y (0)- 2 sY ( s ) + 2 y (0) + 5 Y ( s ) = e L { u 1 ( t ) e t- 1 } = e e- s s- 1 . Now using the initial conditions and leaving Y ( s ) alone on the left: Y ( s ) = e e- s ( s- 1)( s 2- 2 s + 5) = e e- s ( s- 1)(( s- 1) 2 + 4) . We now expand this product into partial fractions so that we obtain a sum: let s- 1 = p then: 1 p ( p 2 + 4) = A p + Bp + C p 2 + 4 ⇒ Ap 2 + 4 A + Bp 2 + Cp = 1 ⇒ A = 1 / 4 , B =- 1 / 4 , C = 0 . Then we have to invert the equality in s-domain: Y ( s ) = e e- s 4 1 s- 1- s- 1 ( s- 1) 2 + 4 | {z } H ( s ) . Let us denote by h ( t ) the inverse Laplace of H ( s ). Then: h ( t ) = e t- e t cos 2 t. Hence y ( t ) = ( e/ 4) L- 1 { e- s H ( s ) } = ( e/ 4) u 1 ( t ) h ( t- 1). Namely: y ( t ) = e 4 u 1 ( t )[ e t- 1- e t- 1 cos 2( t- 1)] = u 1 ( t ) e t 4 [1- cos 2( t- 1)] is the unique solution of the problem....
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202s04mt2 - B U Department of Mathematics Math 202...

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