# Bernoulli Differential Equation

1. Aug 26, 2010

$$\frac{dy}{dx} - \frac{1}{x}*y + \frac{1}{y^2}*x = 0$$

This is a bernoulli differential equation which follows the following procedure
http://en.wikipedia.org/wiki/Bernoulli_differential_equation
Can someone provide the full steps because somewhere I am mistaken?

2. Aug 26, 2010

### JJacquelin

See attachment :

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3. Aug 26, 2010

Thank you a lot. One more question:
In ODE I reach in a place where i have that
$$\frac{dz}{dx}*x^3- 3x^2*z + 3x^4 = 0$$

I cannot find the derivatve on that

Last edited: Aug 26, 2010
4. Aug 26, 2010

### JJacquelin

I cannot understand what you mean : The derivative of what function ?
Well, the equation is
x(dz/dx)-3xz = -3x²
So , solving x(dZ/dx)-3x = 0 leads to Z = C*(x^3)
Remplace C by f(x)
then z = f(x)*(x^3)
dz/dx = (x^3)*(df/dx) +3*x² f
x( (x^3)*(df/dx) +3*x² f ) -3x (x^3)*f = -3x²
(x^4)*(df/dx) = -3x²
df/dx = -3/x²
f = (3/x) + C
z = 3 x² + C (x^3)

5. Aug 26, 2010

x(dz/dx)-3xz = -3x²
how did you find that?

6. Aug 27, 2010

### JJacquelin

Sorry, there was a typo. The correct equation is
x(dz/dx) - 3z = -3x²
and there was other typo in my message #4.
As a consequence, my message #4 is wrong.
Nevertheless, it shows the method to solve this kind of equations.