# Non-exact differential equation where does not exist integrating factor dependent on x and y only

Our calculus professor gave us a few supplementary problems on differential equations and I'm trying to solve the non-exact differential equation $$(6y+x^2y^2)+(8x+x^3y)y'=0$$ I tried finding both $$\frac{M_y-N_x}{N}$$ and $$\frac{M_y-N_x}{-M}$$, but neither give a function only dependent of x or y only.

Any other way to approach this equation?

• Unsure, but I think one of your expressions is wrong (it's been a while since I studied exact DEs, but I think one of the terms should contain an $N_y-M_x$ or a $M_x-N_y$). Jun 5, 2020 at 17:32
• @Cardioid_Ass_22 I checked both Wikipedia and my textbook, I think it is correct. Jun 5, 2020 at 17:34

$$(6y+x^2y^2)dx+(8x+x^3y)dy=0$$ $$(6ydx+8xdy)+(x^2y^2dx+x^3ydy)=0$$ Divide by $$x^2y$$: $$3\dfrac {dx}{x^2}+4\dfrac {dy}{xy}+\dfrac 12d(xy)=0$$ Multiply by $$(xy)^4$$: $$3x^2y^4 {dx}+4x^3y^3dy+\dfrac 12(xy)^4d(xy)=0$$ It's exacy now: $$(y^4 {dx^3}+x^3dy^4)+\dfrac 12(xy)^4d(xy)=0$$ $$d(x^3y^4)+\dfrac 12(xy)^4d(xy)=0$$ Integrate. $$10x^3y^4+(xy)^5=C$$
So the integrating factor $$\mu (x,y)$$ depends on both $$x$$ and $$y$$ and we have: $$\mu (x,y)=x^2y^3$$
• Oh wow, thanks. I was wondering is there any systematic way to find these integrating factors that depends on both $x$ and $y$. Or is it just trial and error? Jun 5, 2020 at 17:54
• You're welcome. Here I tried by trial inspection and error. But you can choose $\mu=x^ay^b$ and make the DE exact. @Rayson Jun 5, 2020 at 17:56