# solutions5ac - Math 33b Quiz 5ac Name UCLA ID 1 Find the...

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Math 33b, Quiz 5ac, February 26, 2008 Name: UCLA ID: 1. Find the solution with initial conditions y (0) = 0, y 0 (0) = 0 to the differential equation y 00 + y = ( 1 , 0 t < 3 e - 2( t - 3) , t 3 . Solution. Observe that the function f ( t ) on the right-hand-side of the differential equation can be written as 1( H ( t ) - H ( t - 3))+ e - 2( t - 3) H ( t - 3) = H ( t )+ H ( t - 3)[ - 1+ e - 2( t - 3) ]. The Laplace transform of this function is 1 s + e - 3 s [ - 1 s + 1 s +2 ]. To solve the differential equation, we take the Laplace transform of both sides of the equation y 00 ( t ) + y ( t ) = f ( t ): ( s 2 Y ( s ) - sy (0) - y 0 (0)) + Y ( s ) = 1 s + e - 3 s [ - 1 s + 1 s + 2 ] Y ( s )( s 2 + 1) = 1 s + e - 3 s [ - 1 s + 1 s + 2 ] Y ( s ) = 1 s 1 s 2 + 1 + e - 3 s [ - 1 s 1 s 2 + 1 + 1 s + 2 1 s 2 + 1 ] We need to write 1 s 1 s 2 +1 and 1 s +2 1 s 2 +1 as sums of partial fractions. First we write 1 s 1 s 2 +1 as A s + Bs + C s 2 +1 . Multiplying by s ( s 2 + 1), we get A ( s 2 + 1) + ( Bs + C )( s ) = 1. From this we get A + B = 0, A = 1, C = 0, so B = - 1 and 1 s 1 s 2 +1 = 1 s - s s 2 +1 . Similarly we write 1 s +2 1 s 2 +1 as D s +2 + Es + F s 2 +1 . Multiplying by s + 2( s 2
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Unformatted text preview: Es + F )( s + 2) = 1. Substituting in s =-2, we have 3 D = 1, so D = 1 3 . Comparing coeﬃcients of s 2 , we have D + E = 0, so E =-1 3 . Comparing constant coeﬃcients, we have D + 2 F = 1, so F = 1 3 . Hence 1 s +2 1 s 2 +1 = 1 3 1 s +2 + 1 3-s +1 s 2 +1 . We now write our Laplace-transformed equation as Y ( s ) = 1 s-s s 2 + 1 + e-3 s [-1 s + s s 2 + 1 + 1 3 1 s + 2 + 1 3-s + 1 s 2 + 1 ] The inverse Laplace transform of the right-hand side is y ( t ) = 1-cos t + H ( t-3)[-1 + cos( t-3) + 1 3 e-2( t-3)-1 3 cos( t-3) + 1 3 sin( t-3)] = 1-cos t + H ( t-3)[-1 + 2 3 cos( t-3) + 1 3 sin( t-3) + 1 3 e-2( t-3) ] . 1...
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