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I have this DEQ: $$3x^2y+2xy+y^3+(x^2+y^2)y^{'}=0$$ and I want to find an integrating factor to make it exact. I derived the formula to do so as $$\mu^{'}=\frac{\mu(M_y-N_x)}{N}$$ I know that $M_y=3x^2+2x, N_x=2x,N=x^2+y^2$, so I can substitute them in the above equation$$\frac{\mu^{'}}{\mu}=\frac{3x^2}{x^2+y^2}$$However, I don't know how to solve for $\mu$ at this point. Can anyone give me a hint on how to proceed from here?

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  • $\begingroup$ Apparently this method isn't apply here. $\endgroup$ – Nosrati Feb 9 '17 at 20:44
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$$(3x^2y+2xy+y^3)dx+(x^2+y^2)dy=0$$ Suppose $F(x,y)$ be an integrating factor so that $\quad (3x^2y+2xy+y^3)F(x,y)dx+(x^2+y^2)F(x,y)dy\quad$ be a total derivative.

If so, the relationship between the partial derivatives must be :

$$\frac{\partial }{\partial y} \big( (3x^2y+2xy+y^3) F(x,y)\big)=\frac{\partial }{\partial x}\big( (x^2+y^2)F(x,y)\big)$$

$$(3x^2+2x+3y^2)F+(3x^2y+2xy+y^3) \frac{\partial F}{\partial y} = (2x)F + (x^2+y^2)\frac{\partial F}{\partial x}$$ After simplification : $$(3x^2y+2xy+y^3) \frac{\partial F}{\partial y} - (x^2+y^2)\frac{\partial F}{\partial x}+3(x^2+y^2)F=0$$ The goal is not to solve this PDE for the general solution, but only to find a particular solution (as simple as possible).

Obviously, if we try $F$ function of $x$ only, the simplification is big : $$\frac{\partial F}{\partial y}=0 \quad\to\quad -\frac{dF}{dx}+3F=0$$ $$F=e^{3x}$$ So, a integrating factor is $\quad e^{3x}$ .

I suppose that you can take it from here to find the general solution of the ODE (on the form of implicit equation): $$e^{3x}\left(y^3+3x^2y\right)=C$$ If you want the solution on explicit form, solve the cubic equation for $y$.

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