# Bernoulli equations with trig

1. Mar 28, 2009

### muso07

1. The problem statement, all variables and given/known data
Find the general solution of the following differential equation.
y'+4xy=10xy2cos(x2)

2. Relevant equations
The usual Bernoulli equation ones:
y'+p(x)y = g(x)ya
u(x)=[y(x)]1-a
u'+(1-a)pu = (1-a)g

3. 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$$\int$$p(x)dx[$$\int$$r(x)e$$\int$$p(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.

Last edited: Mar 29, 2009
2. Mar 29, 2009

### muso07

Nevermind, I figured it out.