q5tts - Solution: Following variation of parameters look...

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Quiz V 1. Find the solution to y ′′ + y = e t + t satisfying y (0) = 1 and y (0) = 0. Solution: A particular solution here will be the sum of solutions for y ′′ + y = e t and y ′′ + y = t . For the inhomogeneous term e t use y = Ae t and find A = 1 / 2. For the inhomogeneous term t use y = At + B , but that leads immediately to A = 1 and B = 0. Since the general solution to the homogeneous equation is y = c 1 cos t + c 2 sin t , you have the general solution y ( t ) = 1 2 e t + t + c 1 cos t + c 2 sin t. The initial condition y (0) = 1 gives 1 / 2+ c 1 = 1 and y (0) = 0 gives 1 / 2+1+ c 2 = 0. So Y ( t ) = 1 2 e t + t + 1 2 cos t - 3 2 sin t is the solution asked for here. 2. The functions y 1 ( t ) = t and y 2 ( t ) = t 2 are solutions to t 2 y ′′ - 2 ty + 2 y = 0. Use them to find a particular solution to t 2 y ′′ - 2 ty + 2 y = t 5 . Be careful using formulas here.
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Unformatted text preview: Solution: Following variation of parameters look for the solution in the form y = c 1 y 1 + c 2 y 2 with the additional condition c ′ 1 y 1 + c ′ 2 y 2 e = 0. That leads to the equations c ′ 1 y 1 + c ′ 2 y 2 e = 0 and c ′ 1 y ′ 1 + c ′ 2 y ′ 2 = t 3 . [Why t 3 and not t 5 ? In standard form this equation is y ′′-(2 /t ) y ′ + (2 /t 2 ) y = t 3 .] The Wronskian W ( y 1 , y 2 ) = t (2 t )-1( t 2 ) = t 2 , and you have c ′ 1 =-y 2 F/W =-t 3 and c ′ 2 = y 1 F/W = t 2 So a particular solution is y =-t 5 / 4 + t 5 / 3 = t 5 / 12....
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This note was uploaded on 02/06/2011 for the course MATH 33a taught by Professor Lee during the Fall '08 term at UCLA.

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