1. ## Solving non-linear ODE

Hi, I would appreciate any help in solving this:

$y''(x) \cdot y(x) = (y'(x))^2 - y'(x)$

With initial conditions $y(0) = 1, y'(0) = 2$

It may be helpful to note the fact that $(y(x) \cdot y'(x))' = (y'(x))^2 + y(x) \cdot y''(x)$

or that $\left( \frac{y'(x)}{y(x)} \right) ' = \frac{y''(x) \cdot y(x) - (y'(x))^2}{(y(x))^2}$ (we get this if we divide both sides of the equations by y^2)

But I was not able to solve this using either.

Thanks for any help.

2. Originally Posted by Defunkt
Hi, I would appreciate any help in solving this:

$y''(x) \cdot y(x) = (y'(x))^2 - y'(x)$

With initial conditions $y(0) = 1, y'(0) = 2$

It may be helpful to note the fact that $(y(x) \cdot y'(x))' = (y'(x))^2 + y(x) \cdot y''(x)$

or that $\left( \frac{y'(x)}{y(x)} \right) ' = \frac{y''(x) \cdot y(x) - (y'(x))^2}{(y(x))^2}$ (we get this if we divide both sides of the equations by y^2)

But I was not able to solve this using either.

Thanks for any help.
The last hint is very useful.

$
\frac{y y'' - y'^2}{y^2} = - \frac{y'}{y^2}
$
or $\left(\frac{y'}{y}\right)' = \left(\frac{1}{y} \right)'$

so

$
\frac{y'}{y} = \frac{1}{y} + c_1
$

or $y' - c_1 y =$1 (this is separable)

3. Originally Posted by Danny
The last hint is very useful.

$
\frac{y y'' - y'^2}{y^2} = - \frac{y'}{y^2}
$
or $\left(\frac{y'}{y}\right)' = \left(\frac{1}{y} \right)'$

so

$
\frac{y'}{y} = \frac{1}{y} + c_1
$

or $y' - c_1 y =$1 (this is separable)
Yep, for some reason I forgot to add the constant to the integral :\

thanks