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

[fluidistic]

Homework Statement

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

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 [itex]y^2[/itex] to get the new DE [itex]\frac{y''}{y}- \left ( \frac{y'}{y} \right ) ^2-6x=0[/itex]. Now I notice that [itex]\left ( \frac{y'}{y} \right )'=\frac{y''}{y}-1[/itex] so that the DE to solve reduces to [itex]\left ( \frac{y'}{y} \right )'- \left ( \frac{y'}{y} \right ) ^2+1-6x=0[/itex].
This suggests me to call a new variable [itex]v=\frac{y'}{y}[/itex]. Thus the DE to solve reduces to [itex]v'-v^2+1-6x=0[/itex]. It is separable so I'm extremely lucky. I reach that [itex]\ln y = \int (3x^2+c_1)dx+c_2 \Rightarrow y(x)=e^{x^3+c_1x}+c_2[/itex].
Hence [itex]y'=(3x^2+c_1)e^{x^3+c_1x}[/itex] and [itex]6xe^{x^3+c_1x}+(3x^2+c_1)^2e^{x^3+c_1x}[/itex]. Plugging these into the original DE doesn't reduces to 0.
What did I do wrong?

 

[SammyS]

[itex]\displaystyle \left(\frac{y'}{y} \right )'=\frac{y''}{y}-(y')^2[/itex]

Added in Edit:

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

[itex]\displaystyle \left(\frac{y'}{y} \right )'=\frac{y''}{y}-\left(\frac{y'}{y}\right)^2[/itex]

 

[fluidistic]

[itex]\displaystyle \left(\frac{y'}{y} \right )'=\frac{y''}{y}-(y')^2[/itex]

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 [itex]y(x)=Ae^{3x^2+c_1x}[/itex] (this work). Thank you very much!