# Bernouli Equation

1. Aug 6, 2009

### n1person

Recently I've been going back in my differential equation book to review some differential equation solving skills, in particular bernouli, ricatti's, and clairaut's equations; simple things enough. However when doing the exercises I have kept running into a "problem" with one question.

$$x \frac{dy}{dx} + y = \frac{1}{y^2}$$

Now I get it into the form to use the bernouli equation:

Divide by x

$$\frac{dy}{dx} + \frac{y}{x} = \frac{1}{xy^2}$$

Multiply by y2

$$y^2 \frac{dy}{dx} + \frac{y^3}{x} = \frac{1}{x}$$

So $$w=y^3$$

$$\frac{dw}{dx}=3y^2\frac{dy}{dx}$$

So now the equation is $$\frac{dw}{dx} + \frac{3w}{x} = \frac{3}{x}$$

So the integrating factor is $$e^\int^\frac{3}{x}^d^x=x^3$$

So you get $$\frac{d(wx^3)}{dx}=3x^2$$

w=1, so $$y^3=w=1=y$$

Which one can see works easily when plugging into the original formula. However the book gives a different answer of $$y^3=1+cx^-^3$$ which I don't understand how one could get with bernouli (and my calculator gives me the same answer with deSolve)... any help?

2. Aug 6, 2009

### rock.freak667

Now if
$$\frac{d(wx^3)}{dx}=3x^2$$

then $wx^3= ?$

When you get that, sub back w=y3 and then divide by x3

3. Aug 7, 2009

### n1person

ahhhh!!! I forgot the constant of integration :( how silly of me....