# Bernouill diff eqn.

1. Aug 6, 2006

### suspenc3

$$y'+\frac{2}{x}y=\frac{y^3}{x^2}$$

$$u=y^-^2$$

so $$y'+(1-3)p(x)u=(1-3)q(x)$$

$$y'-2(2x)u=-2y^3/x^2$$

$$y'-4x/y^2=-2y^3/x^2$$

integration factor is $$e^{{-2x}^2}$$

is this right so far?

2. Aug 6, 2006

### d_leet

I can't follow your work, you make a substitution, but then it seems that you don't really use it because the next to last equation you have involves both u and y when making it should only involve one or the other not both. What are p(x) and q(x) you don't define them and it isn't obvious at all what they are.

When you have a differential equation like this you want to make the substitution that you did, but then you should divide the entire equation by the highest power of y that is present and then use the substitution so that you will get a linear first order differential equation in u, and then you can solve for the function u and use the substitution to find what y should be.

EDIT: Oh I can see what you did now, but none of it is justified at all. When you make that substitution you need to use it. It seems as though you arbitrarily picked out functions to be p(x) and q(x) for no reason at all and just sort of attached u to the equation, you should have realized that this is wrong since you end up with an equation that is entirely different from the one you started with.

Last edited: Aug 6, 2006
3. Aug 6, 2006

### suspenc3

$$p(x)=2/x$$

$$q(x)=1/x^2$$

Ok, yeah there are no examples just the form of a Bernouill Equation so I dont really know what to do.

I dont see what the point of the sub $$u=y^{1-n}$$ is?

I know it is suppose to make it linear..I gueess mine is wrong though..that might be the reason haha
EDIT:wiki explains it better, im gonna go try it again

Last edited: Aug 6, 2006
4. Aug 6, 2006

### d_leet

Well you didn't really use the substitution initially you kind of just attached a u to the equation so thats why it didn't work.

5. Aug 6, 2006

### suspenc3

ok i tried again let $$u=1/y^2$$
$$y'y^{-3}+ \frac{2}{x}y^{-2} = \frac{1}{x^2}$$

$$du/dx - 4u/x = -2/x^2$$

$$I(x)=e^{2 \int lnx} = x^2$$

$$du/dx x^2 - 4xu = -2$$

$$x^2u = \int-2$$

$$x^2u=-2x+C$$

$$u = -2/x + C/x^2$$

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

Last edited: Aug 6, 2006
6. Aug 6, 2006

### d_leet

Your integrating factor is wrong. You should have integrated -4/x not 2/x. and then raised e to that power to get your integrating factor.

7. Aug 6, 2006

### suspenc3

Riiiiight, Thanks!

Last edited: Aug 6, 2006