# Thread: First Order ODE

1. ## First Order ODE

This one gave me some trouble:

2y(x^2-y+x)dx + (x^2-2y)dy=0

I believe the integrating factor is e^(2x) but I may be wrong.

Thanks on advance, sorry about lack of latex, I'll redo it when I get home and am not on my phone.

2. I was bored so here is more than just the verification of the integrating factor.
$\displaystyle {\partial M\over\partial y}=2x^{2}-4y+2x$

$\displaystyle {\partial N\over\partial x}=2x$

$\displaystyle \frac{{M}_{y}-{N}_{X}}{N(x,y)}=\frac{2x^{2}-4y+2x-2x}{x^{2}-2y}=2$

$\displaystyle {e}^{2\int dx}={e}^{2x}$

$\displaystyle (2yx^{2}{e}^{2x}-2y^{2}{e}^{2x}+2yx{e}^{2x})dx+(x^{2}{e}^{2x}-2y{e}^{2x})dy$

$\displaystyle f(x,y)=\int (x^{2}{e}^{2x}-2y{e}^{2x})dy=x^{2}{e}^{2x}y-y^{2}{e}^{2x}+g(x)$

$\displaystyle x^{2}{e}^{2x}y-y^{2}{e}^{2x}+g(x)$

$\displaystyle {\frac{d}{dx}}[x^{2}{e}^{2x}y-y^{2}{e}^{2x}+g(x)]=2yx^{2}{e}^{2x}-2y^{2}{e}^{2x}+2yx{e}^{2x}+g'(x)$

$\displaystyle g'(x)=0$

$\displaystyle g(x)=C$

$\displaystyle x^{2}{e}^{2x}y-y^{2}{e}^{2x}+C$

3. That's exactly what I wanted, thanks a ton.