Nonlinear Second Order Differential Equations: Solving yy''-(y')^2-6xy^2=0

[fluidistic]

Homework Statement

I must solve yy''-(y')^2-6xy^2=0.

Homework Equations

Not sure.

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?

 

[SammyS]

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

Added in Edit:

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

 

[fluidistic]

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

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!