Differential equation $ydx+(2x-ye^y)dy=0$ has an integrating factor $\mu(x,y)=x^my^n$ for constants $m$ and $n$. Determine $\mu(x,y)$.
I tried to solve using the formula $M\frac {\partial\mu}{\partial y}-N\frac {\partial\mu}{\partial x}=(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})\mu$. But still I didn't managed to find the $\mu(x,y)$.
The answer says $\mu(x,y)=y$.
But I want to know the steps to find this. I can solve the differential equation with $\mu(x,y)=y$. So please help me for finding the integrating factor $\mu(x,y)$.
Best Answer
Another method.
Since you know the integrating factor $\mu$ is of the form $x^{m}y^{n}$, multiply the differential equation by $x^{m}y^{n}$ to obtain $$x^{m}y^{n+1}dx+(2y^{n}x^{m+1}-x^{m}y^{n+1}e^{y})dy=0$$
For the differential equation to be exact we want $$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$
that is $$(n+1)y^{n}x^{m}=2(m+1)y^{n}x^{m}-mx^{m-1}y^{n+1}e^{y}$$
Since there are no $e^{y}$ terms on the LHS we choose $m=0$ which gives $$(n+1)y^{n}x^{0}=2y^{n}x^{0}$$ $$(n+1)y^{n}=2y^{n}$$
Thus $n=1$ and we have $\mu(x,y)=y$ as required.
See this also.