Find general solution, 1st order ODE

[2h2o]

Homework Statement

Find a general solution.

Homework Equations

2x\frac{dy}{dx}+y^{3}e^{-2x}=2xy

The Attempt at a Solution

Looks like a Bernoulli equation to me, after some algebra:

\frac{dy}{dx}+\frac{y^{3}}{2xe^{2x}}=y

\frac{dy}{dx}+\frac{y}{2xe^{2x}}=y^{-1}

so with n=-1
v=y^{2}, y=v^{1/2}, \frac{dy}{dx}=\frac{1}{2}v^{-1/2}\frac{dv}{dx}

\frac{1}{2}v^{-1/2}}\frac{dv}{dx}+\frac{v^{1/2}}{2xe^{2x}}=v^{-1/2}

\frac{1}{2}\frac{dv}{dx}+\frac{v}{2xe^{2x}}=1

\frac{dv}{dx}+\frac{v}{xe^{2x}}=2

Now an integrating factor:

\mu=exp[\int{x^{-1}e^{-2x}dx}]

And that's where I get stuck. This doesn't look like any elementary integral I've learned how to solve, and wolfram|alpha gives me something called the "exponential integral" which we haven't been taught. So I've done something wrong, but I don't see it.

Thanks for any insights.

 

[2h2o]

Ok, I think I found my error: I forgot to divide the dy/dx by y^2 when I obtained y^-1 as the RHS. No wonder I wound up with a nonelementary integral. However, I'm still stuck.

I don't recognize this equation, and manipulating it around trying to "force it" to be in first-order linear, homogeneous, or bernoulli hasn't helped. So I'm stuck.

This is where I am, and don't recognize it:

\frac{dy}{dx}+\frac{y^{3}}{2xe^{2x}}=y

 

[adoado]

Isn't that separable? Split up the differentials?

 

[hunt_mat]

This is just another Bernoulli type ODE isn't it?