Solving Differential EquationsDate: 5/8/96 at 20:59:31 From: Anonymous Subject: Solving Differential Equations I am trying to solve the following differential equation: (2x^2 + 2y^2 - y)dx + (x^2y + y^3 +x)dy = 0 Please help! Thanking you in advance, Alnoor Date: 11/14/96 at 12:26:33 From: Doctor Jimbo Subject: Re: Solving Differential Equations Alnoor, I referred this question to my office-mate, Bernard, who loves to solve differential equations. Here's what he had to say: 1) The equation is not exact and has no easy integrating factor. 2) Use the substitution x = r cos t, y = r sin t, motivated by the occurrence of several x^2+y^2 factors. Regard r as a function of t. 3) In these new variables, the equation is exact. It reads: (2 cos t+r sin^2 t) dr + (-2r sin t+1+r^2 sin t cos t) dt = 0. 4) The implicit solution is: 2 r cos t+r^2/2 sin^2 t+t = c, where c is an arbitrary constant. This can be solved for r = r(t). Then x and y can be recovered by using x = r(t) cos t, y = r(t) sin t, giving a parametric representation of the solution (actually 2 solutions, since there is a plus/minus choice). Hope that helps. -Doctor Jimbo, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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