Find the integrating factor of form $x^m y^n$ of first order ODE to make it exact

Question

For the first-order ODE: $$(2y^2-3xy)dx + (4xy-3x^2)dy = 0,$$

find the integrating factor (of form below) that makes the ODE exact… $$\mu(x,y)=x^my^n$$

My attempt:

$$M_y = 4y – 3x; N_x = 4y – 6x$$

I'm pretty much stuck at this point…any help?

Thanks

Answer 1

$$\frac{\partial}{\partial y}\left(\left(2y^2-3xy\right)x^my^n\right)=x^{m-1}y^{n-1}\left((4+2n)xy^2-(3+3n)x^2y\right) $$ $$\frac{\partial}{\partial x}\left(\left(4xy-3x^2\right)x^my^n\right)=x^{m-1}y^{n-1}\left((4+4m)xy^2-(3m+6)x^2y\right)$$ Since we want $M_y=N_x$, we have to equate the corresponding coefficients: $$\begin{cases} 4+4m=4+2n \\ 3m+6=3+3n \end{cases} $$ The solution is $m=1, n=2$, so the integrating factor is $\mu(x,y)=xy^2$. After you multiply by it, the differential equation is exact, and can be solved in a standard way. The answer is $$x^2y^4-x^3y^3=C $$