# Math Help - integrating factor proof/ partial derivative wrt xy

1. ## integrating factor proof/ partial derivative wrt xy

The full question is:
show that if $\frac{(N_{x} - M_{y})}{(x*M-y*N)}=R$ where R depends only on xy, then the differential equation M+N*y'=0 has an integrating factor of the form U(xy). Find a general formula for this integrating factor.

Using an equation from the text for U (for which finding a solution, U, means that the differential equation above is exact) I have:
$U= \frac{N*U_{x}-M*U_{y}}{M_{y}-N_{x}}$
then substituting in
$M_{y}-N_{x}=(yN-xM)*R$
and equating coefficients of N and M respectively,
$y*U*R=U_{x}$ and $x*U*R=U_{y}$

Is there some way that I can find the derivative of U wrt xy using the partial derivatives of U wrt x and y that I have above? Then I could integrate wrt xy to find U, right?

2. $\mu(x,y)M(x,y)dx+\mu(x,y)N(x,y)\,dx=0.$

we have $\mu(x,y)=h(xy),$ and by the exact condition we have $\displaystyle\frac{\partial \mu }{\partial y}M-\frac{\partial \mu }{\partial x}N=\left( \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y} \right)\mu ,$ now $\mu_x=h'(xy)y$ and $\mu_y=h'(xy)x,$ so $\displaystyle\frac{1}{xM-yN}\left( \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y} \right)=\frac{h'(xy)}{h(xy)}=R(xy),$ put $z=xy$ and solve that little ODE and you got your integrating factor.