Solving non-linear ODE

**Defunkt** · March 8th 2010, 01:05 PM

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.

**Danny** · March 8th 2010, 02:54 PM

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)

**Defunkt** · March 8th 2010, 03:13 PM

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

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