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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.
 Fall '08
 lee

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