sol3 - MAP2302/2331 Exam3 Dr Sin No Calculators Answer the...

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MAP2302/2331 Exam3 Dr Sin No Calculators. Answer the questions in the spaces provided on the question sheets. Please write your answers in full detail. If you run out of room for an answer, continue on the back of the page. Name: 1. (7 points) Let L{ y ( t ) } ( s ) = Y ( s ). Find L{ e 3 t ty ( t ) } ( s ) . Be sure to explain clearly which general properties of the Laplace transform you use. Solution: By the t differentiation formula (5), we have L{ y ( t ) } ( s ) = sY ( s ) - y (0) . Then by the s differentiation formula (6), L{ ty ( t ) } ( s ) = - d ds ( sY ( s ) - y (0)) = - sY ( s ) - Y ( s ) . Finally, by translation in t , L{ e 3 t ty ( t ) } ( s ) = - ( s - 3) Y ( s - 3) - Y ( s - 3) . 2. (8 points) Find the Laplace transform F ( s ) of the function f ( t ) = cos 3 t, if 0 < t < π 0 , if π < t . Solution: We can write use the step function to write f ( t ) = cos 3 t - u ( t - π ) cos 3 t. To take Laplace transforms, we apply formula (4) to the second term, with g ( t ) = cos 3 t and a = π . To use (4), we need to find g ( t + a ). We have g ( t + a ) = cos(3( t + π )) = - cos 3 t . We can now compute F ( s )) = s s 2 + 9 - ( e - πs - s s 2 + 9 ) = s (1 + e - πs ) s 2 + 9 .
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MAP2302/2331 Exam3 Dr Sin 3. (7 points) Solve the IVP y + y = δ ( t - π ) , y (0) = 0 , y (0) = 0 . Solution: Taking Laplace transforms, applying the formula for derivatives yields S 2 Y + Y = e - πs . Hence Y = e - πs . 1 s 2 + 1 . By the translation in s formula (3), we get
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