# Non linear second order DE

1. Oct 9, 2011

### fluidistic

1. The problem statement, all variables and given/known data
I must solve $yy''-(y')^2-6xy^2=0$.

2. Relevant equations
Not sure.

3. The attempt at a solution
I reach something but this doesn't satisfy the original DE...
Here is my work:
I divide the DE by $y^2$ to get the new DE $\frac{y''}{y}- \left ( \frac{y'}{y} \right ) ^2-6x=0$. Now I notice that $\left ( \frac{y'}{y} \right )'=\frac{y''}{y}-1$ so that the DE to solve reduces to $\left ( \frac{y'}{y} \right )'- \left ( \frac{y'}{y} \right ) ^2+1-6x=0$.
This suggests me to call a new variable $v=\frac{y'}{y}$. Thus the DE to solve reduces to $v'-v^2+1-6x=0$. It is separable so I'm extremely lucky. I reach that $\ln y = \int (3x^2+c_1)dx+c_2 \Rightarrow y(x)=e^{x^3+c_1x}+c_2$.
Hence $y'=(3x^2+c_1)e^{x^3+c_1x}$ and $6xe^{x^3+c_1x}+(3x^2+c_1)^2e^{x^3+c_1x}$. Plugging these into the original DE doesn't reduces to 0.
What did I do wrong?

2. Oct 9, 2011

### SammyS

Staff Emeritus
$\displaystyle \left(\frac{y'}{y} \right )'=\frac{y''}{y}-(y')^2$

As pointed out by fluiistic in the following post: The above is wrong. It should be the following.

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

Last edited: Oct 9, 2011
3. Oct 9, 2011

### fluidistic

Ah I see! Thanks for pointing this to me. I think you forgot to divide by y^2 the second term.

Edit: I now reach the result, namely $y(x)=Ae^{3x^2+c_1x}$ (this work). Thank you very much!

Last edited: Oct 9, 2011