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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?

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  • $\begingroup$ 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 $M_x-N_y$). $\endgroup$
    – CardioidAss22
    Jun 5, 2020 at 17:32
  • $\begingroup$ @Cardioid_Ass_22 I checked both Wikipedia and my textbook, I think it is correct. $\endgroup$
    – Rayson
    Jun 5, 2020 at 17:34

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

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  • $\begingroup$ 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? $\endgroup$
    – Rayson
    Jun 5, 2020 at 17:54
  • $\begingroup$ 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 $\endgroup$
    – user577215664
    Jun 5, 2020 at 17:56
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    $\begingroup$ Ok, I'll keep in mind. Thanks a lot.. @Aryadeva $\endgroup$
    – Rayson
    Jun 5, 2020 at 18:01
  • $\begingroup$ You're welcome @Rayson $\endgroup$
    – user577215664
    Jun 5, 2020 at 18:01

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