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### Solving Differential Equations

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Date: 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

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/
```
Associated Topics:
High School Calculus

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