# Differential eq

1. Jan 19, 2013

### aaronfue

1. The problem statement, all variables and given/known data

Given: (x2-y2)dx + (x2-xy)dy=0,

Verify that the following function is a solution for the given differential equation:

c1(x+y)2=xey/x

2. The attempt at a solution

I've gotten this far:

1st - I solved for $\frac{dy}{dx}$ in the given equation.

$\frac{dy}{dx}$=$\frac{-x^2-y^2}{x^2-xy}$

2nd - I used implicit differentiation on the function and got:

2c1(x+y)(1+y)=xey/x($\frac{xy'-y}{x}$) + ey/x

Now....I believe that I can solve for y' in: 2c1(x+y)(1+y)=xey/x($\frac{xy'-y}{x}$) + ey/x?

2. Jan 19, 2013

### ehild

Your method is correct, but there is a sign error in the equation in red, it should be

$\frac{dy}{dx}$=$\frac{-x^2+y^2}{x^2-xy}$

and there are mistakes also in the blue equation:

2c1(x+y)(1+y')=xey/x($\frac{xy'-y}{x^2}$) + ey/x

ehild

3. Jan 19, 2013

### aaronfue

My mistake. In the original equation it was supposed to be (x2 + y2)dx. So my original $\frac{dy}{dx}$=$\frac{-x^2-y^2}{x^2-xy}$ equation was okay. And you are correct about the blue equation. I completely forgot the y'.

Thanks for catching that.

Last edited: Jan 19, 2013