General Solution for a Bernoulli Equation with Trigonometric Functions

[muso07]

Homework Statement

Find the general solution of the following differential equation.
y'+4xy=10xy²cos(x²)

Homework Equations

The usual Bernoulli equation ones:
y'+p(x)y = g(x)y^a
u(x)=[y(x)]^1-a
u'+(1-a)pu = (1-a)g

The Attempt at a Solution

I got up until the general solution part.. I'll just type bits of it out because it'll take me ages (I'm a slow typer).

So in the equation, a=2, p(x)=4x, g(x)=10xcos(x²)

Change of variables:
u(x)=[y(x)]^1-a = y^-1

and u'+(1-a)pu = (1-a)g
=> u'-4xu = -10xcos(x²)

Now here's where I get confused..
General solution:
u= e^\intp(x)dx[\intr(x)e^\intp(x)dxdx+c
where p=-4x, r=-10xcos(x²)

u= e^{\int-2x^2}[\int(-10xcos(x²)e^{\int-2x^2})dx+c

Argh.. that's messy, I hope it makes sense.

Anyway, I can't seem to figure out that integral... I've used parts and stuff but I don't seem to be getting anywhere.

Any help would be appreciated.

 

[muso07]

Nevermind, I figured it out.