# Implicit Integrating Factor

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?

$$y - 3y^3 = \left(y^4 + 2x\right)\frac{dy}{dx}$$ 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.
• I knew there was something wrong with treating $y$ as a constant. Thanks so much!
• 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 $\frac{1}{y^3}$ . Commented Feb 3, 2019 at 6:15