# Ugh Bernoulli Equations

1. Sep 13, 2009

### darthxepher

Ugh Bernoulli Equations!!!

1. The problem statement, all variables and given/known data

dy/dx = y(xy^3-1)

I tried to set it up and use the bernoulli equation method as a substitution but it didn't work. Any tips?

3. The attempt at a solution

I set u=y^(-3) and had it set up like this

dy/dx + y = xy^4

Ok HELP!

Thanks...

2. Sep 13, 2009

### Gregg

Re: Ugh Bernoulli Equations!!!

I've never done this before, but:

$$\frac{dy}{dx} = f(x)y + g(x)y^k$$

The solution is

$$y^{1-k} = y_1 + y_2$$

Where

$$y_1= Ce^{\phi(x)}$$

$$y_2=(1-k)e^{\phi(x)}\int e^{-\phi(x)}g(x)dx$$

$$\phi(x) = (1-k)\int f(x)dx$$

Here,

$$k=4$$

$$g(x)=x$$

$$f(x)=-1$$

So,

$$\phi(x) = 3x$$

Then,

$$y_1= Ce^{3x}$$

$$y_2=-3e^{3x}\int xe^{-3x}dx=\frac{1}{3}+x$$

Finally,

$$y = (Ce^{3x} + \frac{1}{3} + x)^{-\frac{1}{3}}$$

I think...

3. Sep 13, 2009

### LCKurtz

Re: Ugh Bernoulli Equations!!!

OK, writing it as y' + y = xy^4 is a good start. What you want to do next is divide by y^4:

y^(-4) y' + y^(-3) = x. Now let v = y^(-3) so v' = -3y^(-4) y'

This gives you expressions for y^(-4)y' and y^(-3) in terms of v and it gives you a first order linear equation in v which you can solve. Then back substitute to get y.