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Unformatted text preview: B U Department of Mathematics Math 202 Differential Equations Spring 2005 Second Midterm This archive is a property of Bo˘gazi¸ ci University Mathematics Department. The purpose of this archive is to organise and centralise the distribution of the exam questions and their solutions. This archive is a non-profit service and it must remain so. Do not let anyone sell and do not buy this archive, or any portion of it. Reproduction or distribution of this archive, or any portion of it, without non-profit purpose may result in severe civil and criminal penalties. 1. Use variation of parameters to find the general solution to x 2 y 00 + 3 xy + y = 1 x for x > . Solution: The corresponding homogenous equation x 2 y 00 + 3 xy + y = 0 is Euler’s equation. The indicial equation is r 2 + (3- 1) r + 1 = r 2 + 2 r + 1 = ( r + 1) 2 = 0 . So we have equal roots r 1 = r 2 =- 1. Hence, y 1 ( x ) = x- 1 , y 2 ( x ) = x- 1 ln x form a fundamental set of solutions. Now by variation of parameters, we assume y ( x ) = c 1 ( x ) y 1 + c 2 ( x ) y 2 is a particular solution. We insert y in the given differential equation. Assuming, as usual, that c 1 y 1 + c 2 y 2 = 0 we obtain as the second identity: x 2 ( c 1 y 1 + c 2 y 2 ) = 1 x ; that is, we get: y 1 y 2 y 1 y 2 c 1 c 2 = x...
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