If we have a non-exact ODE, then to convert it to an exact ODE we multiply the ODE with an integrating factor $\mu(x,y)$.
Lets us say we have the following ODE: $$M(x,y)dx+N(x,y)dy=0,$$ and let us denote $\frac{\partial M(x,y)}{\partial y}=M_y$ and $\frac{\partial N(x,y)}{\partial x}=N_x$.
Since the ODE is not exact, $M_y-M_x =f(x,y)\neq 0$.
We know that if $\frac{f(x,y)}{N(x,y)}$ depends only on $x$ then $e^{\int\frac{f(x,y)}{N(x,y)}dx}$ is the integrating factor of the ODE and if $\frac{f(x,y)}{M(x,y)}$ only depends on $y,$ then $e^{\int\frac{f(x,y)}{-M(x,y)}dy}$ is the integrating factor.
But what is the integrating factor of the ODE when both $\frac{f(x,y)}{M(x,y)}$ and $\frac{f(x,y)}{N(x,y)}$ are functions of $x$ and $y$ and neither of them are independent of any variable?
For example, in the following ODE: $$y(1+2x^2+2y^2)\; \mathrm dx + x(1-2x^2-2y^2)\; \mathrm dy=0,$$ $M_y-N_x=f(x,y)=8x^2+8y^2$ and neither of $\frac{f(x,y)}{M(x,y)}$ and $\frac{f(x,y)}{N(x,y)}$ are independent of any variable i.e both of them depend on $x$ and $y$ both.
So in such cases how do we determine the integrating factor?
Best Answer
I believe there is no general procedure to find integrating factors in every case. There are some special cases where it is possible as you commented. I'm gonna point out another one:
If $$ \frac{N_x-M_y}{xM-yN} = g(xy) $$ for some function $g$ of one variable (here $g(xy)$ means that the function $g$ is evaluated in the number $xy$. I mention this because often people think that I missed a "$,$" separating the varaibles), and if $f$ is a primitive of $g$, meaning that $f'(t)=g(t)$ for all $t$, then $\mu=e^{f(xy)}$ is an integrating factor (you can prove it by yourself).
So, in this case, you obtain $$ \frac{N_x-M_y}{xM-yN} = -\frac{2}{xy}, $$ and we can take $g(t)=-2/t$. A primitive for $g$ is given by $f(t)=-2\ln|t|=\ln (t^{-2})$, and hence $$ \mu = e^{\ln((xy)^{-2})} = \frac{1}{x^2y^2} $$ is an integrating factor.