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# solutions5bd - Math 33b Quiz 5bd Name UCLA ID 1 Find the...

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Math 33b, Quiz 5bd, February 28, 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 = ( t, 0 t < 1 cos 2( t - 1) , t 1 . (A small table of Laplace transforms is on the back of the quiz.) Solution. Observe that the function f ( t ) on the right-hand-side of the differential equation can be written as t ( H ( t ) - H ( t - 1)) + cos 2( t - 1) H ( t - 1) = tH ( t ) + H ( t - 1)[ - t + cos 2( t - 1)] = tH ( t ) + H ( t - 1)[ - ( t - 1) - 1 + cos 2( t - 1)] . The Laplace transform of this function is 1 s 2 + e - s [ - 1 s 2 - 1 s + s s 2 +4 ]. 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 2 + e - s [ - 1 s 2 - 1 s + s s 2 + 4 ] Y ( s )( s 2 + 1) = 1 s 2 + e - s [ - 1 s 2 - 1 s + s s 2 + 4 ] Y ( s ) = 1 s 2 1 s 2 + 1 + e - s [ - 1 s 2 1 s 2 + 1 - 1 s 1 s 2 + 1 + s s 2 + 4 1 s 2 + 1 ] We need to write 1 s 2 1 s 2 +1 , 1 s 1 s 2 +1 , and s ( s 2 +1)( s 2 +4) as sums of partial fractions. First we write 1 s 2 1 s 2 +1 as A s + B s 2 + Cs + D s 2 +1 . Clearing denominators, we get 1 = As ( s 2 + 1) + B ( s 2 + 1) + ( Cs + D )( s 2 ). Substituting s = 0, we get B = 1. Checking the coefficients of s 2 , we get B + D = 0, so D = - 1. Checking coefficients of s , we get A = 0. Checking coefficients of s 3 , we get A + C = 0, so C = 0. Hence 1 s 2 1 s 2 +1 = 1 s 2 - 1 s 2 +1 . (This is also easy to see by inspection.) Now we write
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