# integrating factor proof/ partial derivative wrt xy

• Oct 3rd 2010, 08:58 PM
234578
integrating factor proof/ partial derivative wrt xy
The full question is:
show that if $\displaystyle \frac{(N_{x} - M_{y})}{(x*M-y*N)}=R$ 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:
$\displaystyle U= \frac{N*U_{x}-M*U_{y}}{M_{y}-N_{x}}$
then substituting in
$\displaystyle M_{y}-N_{x}=(yN-xM)*R$
and equating coefficients of N and M respectively,
$\displaystyle y*U*R=U_{x}$ and $\displaystyle x*U*R=U_{y}$

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?
• Oct 4th 2010, 05:57 AM
Krizalid
$\displaystyle \mu(x,y)M(x,y)dx+\mu(x,y)N(x,y)\,dx=0.$

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