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Unformatted text preview: B U Department of Mathematics Math 202 Differential Equations Spring 2005 First 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. Consider the differential equation ( xy- 1) dx + ( x 2- xy ) dy = 0. (a) Show that this is not an exact equation. (b) Find an integrating factor μ to make it exact. (c) Multiply the equation by μ on both sides and solve the differential equation. (d) State conditions on x and y required to make your solution valid. Solution: Let M ( x, y ) = xy- 1 and N ( x, y ) = x 2- xy . Then ∂ ∂y M = x ; ∂ ∂x N = 2 x- y . Since M y 6 = N x , the DE is not exact. We find μ ( x, y ) such that ∂ ∂y ( μM ) = ∂ ∂x ( μN ). We require: μ y ( xy- 1) + xμ = μ x ( x 2- xy ) + (2 x- y ) μ. Assuming that μ = μ ( x ) i.e. μ y = 0 we get:- xμ x = μ which is in accordance with our assumption. Hence we get as an integrating factor, for example, μ =...
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