MAT 275 Practice Final Exam A Key.pdf - MAT 275 Practice...

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MAT 275: Practice Final Exam A - Key Author: Blake Marx Solving IVP’s with Discontinuous Forcing Functions: 1. Find the y ( t ) that satisfies y 00 + y = f ( t ) = t 0 t < 1 3 + 2 t 1 t < 3 t 2 t 3 , y (0) = - 1 , y 0 (0) = 3 . Write f ( t ) = t + (3 + t ) u 1 ( t ) + ( t 2 - 2 t - 3) u 3 ( t ). In order to shift 3 + t, t 2 - 2 t - 3, compute the Taylor series centered at 1 and 3 respectively. We see that: 3 + t = 4 + ( t - 1) , t 2 - 2 t - 3 = 4( t - 3) + ( t - 3) 2 So: f ( t ) = t + [4 + ( t - 1)] u 1 ( t ) + [4( t - 3) + ( t - 3) 2 ] u 3 ( t ) Taking the Laplace Transform of the differential equation ( Y = L { y } ): L y 00 + y = t + [4 + ( t - 1)] u 1 ( t ) + [4( t - 3) + ( t - 3) 2 ] u 3 ( t ) = s 2 Y - sy (0) - y 0 (0) + Y = 1 s 2 + [ 4 s + 1 s 2 ] e - s + [ 4 s 2 + 2 s 3 ] e - 3 s = ( s 2 + 1) Y = 1 s 2 + [ 4 s + 1 s 2 ] e - s + [ 4 s 2 + 2 s 3 ] e - 3 s - s + 3 = Y = 1 s 2 ( s 2 +1) + [ 4 s ( s 2 +1) + 1 s 2 ( s 2 +1) ] e - s + [ 4 s 2 ( s 2 +1) + 2 s 3 ( s 2 +1) ] e - 3 s - s s 2 +1 + 3 s 2 +1 By Partial Fraction Decomposition: 1 s ( s 2 +1) = 1 s - s s 2 +1 , 1 s 2 ( s 2 +1) = 1 s 2 - 1 s 2 +1 , 1 s 3 ( s 2 +1) = 1 s 3 - s s 2 +1 - 1 s : Y = 1 s 2 - 1 s 2 +1 + [ 4 s - 4 s s 2 +1 + 1 s 2 - 1 s 2 +1 ] e - s + [ 4 s 2 - 4 s 2 +1 + 2 s 3 - 2 s s 2 +1 - 2 s ] e - 3 s - s s 2 +1 + 3 s 2 +1 = y ( t ) = L - 1 { 1 s 2 - 1 s 2 +1 +[ 4 s - 4 s s 2 +1 + 1 s 2 - 1 s 2 +1 ] e - s +[ 4 s 2 - 4 s 2 +1 + 2 s 3 - 2 s s 2 +1 - 2 s ] e - 3 s - s s 2 +1 + 3 s 2 +1 } = y ( t ) = t - sin( t ) + [4 - 4 cos( t - 1) + ( t - 1) - sin( t - 1)] u 1 ( t ) +[4( t - 3) - 4 sin( t - 3) + ( t - 3) 2 - 2 cos( t - 3) - 2] u 3 ( t ) - cos( t ) + 3 sin( t ) Simplifying: y ( t ) = t +2 sin( t ) - cos( t )+[ t +3 - 4 cos( t - 1) - sin( t - 1)] u 1 ( t )+[ t 2 - 2 t - 5 - 4 sin( t - 3) - 2 cos( t - 3)] u 3 ( t ) Impulse Functions: 1. Find the y ( t ) that satisfies y 00 + 9 y = 2 δ ( t - 3), y (0) = 1 , y 0 (0) = 2 .

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