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?
$$(6y+x^2y^2)dx+(8x+x^3y)dy=0$$ $$(6ydx+8xdy)+(x^2y^2dx+x^3ydy)=0$$ 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: $$\mu (x,y)=x^2y^3$$