# Differential equation

1. Sep 14, 2009

### intenzxboi

Give the general solution of:

dy/dx = 6x y^2 -6x + 3 y^2 -3.

No clue how to start. Guessing you would use Bernoulli's equation but i don't see how it is possible to put in the y'(x) + p(x)y(x) = q(x) y(x)^n form.

2. Sep 14, 2009

### Gregg

Bernoulli Equation

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

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

where

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

$$y_1 = Ce^\phi$$

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

3. Sep 14, 2009

### Gregg

Actually don't bother, you can rearrange to get this:

$$\frac{dy}{dx} = (6x+3)(y^2-1)$$

you can take it from there, no?

4. Sep 14, 2009

### intenzxboi

Yes i can thank you

5. Sep 14, 2009

### intenzxboi

actually im still having trouble do i still use the bernoulli method?

so it is a simple separation problem?

6. Sep 14, 2009

### Gregg

Sorry I didn't realise you replied.

$$\frac{dy}{dx} = (6x+3)(y^2-1)$$

$$\int \frac{dy}{y^2-1} = \int (6x+3) dx$$

$$\frac{1}{2}log\left(\frac{1-y}{1+y}\right) = 3x^2+3x + C$$

$$\sqrt{\frac{1-y}{1+y}} = Ae^{3x^2+3x}$$

$$y=\frac{1-Ae^{6x^2+6x}}{1+Ae^{6x^2+6x}}$$

Something like this I think?

Last edited: Sep 14, 2009
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