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I have the ODE given below and am told to first find a suitable integrating factor to obtain an implicit solution $F(x,y)=C$ and then solving explicitly for $x$.

$y - 3y^3 = \left(y^4 + 2x\right)y'$ , $y(0)=1$

I shuffled the terms around to get:

$\frac{dy}{dx}+y(\frac{3y^2-1}{y^4+2x})=0$

Then found the integrating factor by doing

$u=e^{^\int\frac{3y^2-1}{y^4+2x}}=e^\frac{(3y^2-1)(ln|y^4+2x|)}{2}$

$e^\frac{(3y^2-1)(ln|y^4+2x|)}{2}y=C$

Then taking $ln$ of both sides and shuffling terms around to finally get:

$x=\frac{e^\frac{2ln|y|}{1-3y^2}-y^4}{2}$

However, this answer is wrong. Is my method wrong?

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$$y - 3y^3 = \left(y^4 + 2x\right)\frac{dy}{dx}$$ Your calculus is wrong. You wrote : $$u=e^{\int\frac{3y^2-1}{y^4+2x}}$$ which is non-sens because it is not specified to which variable $x$ , or $y$ , or both related, the integration has to be carried out.

Note that $\int\frac{3y^2-1}{y^4+2x}dx \neq \frac{(3y^2-1)(ln|y^4+2x|)}{2}$ because $y(x)$ is not constant.

A more direct approach is to consider the inverse function $x(y)$ . The ODE becomes : $$(y - 3y^3)x'-2x = y^4 $$ This is a first order linear ODE that you can solve with the standard method.

If you don't see it at first, change of variables : $X=y$ and $Y=x$ which leads to $$(X-3X^3)=(X^4+2Y)\frac{dX}{dY} \quad\to\quad (X-3X^3)\frac{dY}{dX}-2Y=X^4$$ The solution is : $$x=\frac{y^4}{2-6y^2}+C\frac{y^2}{1-3y^2}$$ I am sure that you can take it from here.

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  • $\begingroup$ I knew there was something wrong with treating $y$ as a constant. Thanks so much! $\endgroup$
    – Sami
    Commented Feb 3, 2019 at 5:40
  • $\begingroup$ Note that if you want to solve $(y-3y^3)dx-(y^4+2x)dy=0$ with the method of integrating factor, the integrating factor is $\frac{1}{y^3}$ . $\endgroup$
    – JJacquelin
    Commented Feb 3, 2019 at 6:15

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