Nonlinear Ordinary Differential Equation Help

lilyrose
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Homework Statement



y'=(x^2 +xy-y)/((x^2(y)) -2x^2)[/B]

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The Attempt at a Solution



I know that really the only way to solve this one is to use an integrating factor, and make it into an exact equation. My DE teacher said that to make it into a exact equation you need to take the partial of P(x,y) with respect to y minus the partial of (x,y) with respect to x, all divided by Q(x,y). However, when I do this, I get a nasty equation: (x^2 +5x-1-2xy)/(x^2 -2x^2).
We were told that you can only find μ, the integrating factor, if we got an equation in terms of one variable. Where am I going wrong?[/B]
 
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I would start by writing this equation as (x^2+ xy- y)dx+ (2x^2- x^2y)dy= 0. Now, the "P" and "Q" your teacher was talking about are P= x^2+ xy- y and Q= 2x^2- x^2y= x^2(2- y). The partial derivative of P with respect to y is P_y= x- 1 and the partial derivative of Q with respect to x is 2x(2- y)= 4x- xy. Their difference is x- 1- (4x- xy)= xy+ 5x- 1. Dividing that by Q, \frac{xy+ 5x- 1}{x^2(2- y)}, almost what you have. That is not an equation, it is an expression. What are you to do with it?

We were told that you can only find μ, the integrating factor, if we got an equation in terms of one variable.
you must have misunderstood. That is certainly not true.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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