# DE Initial Value Problem

I'm given the following DE and initial conditions:
$$y''=2yy'$$
$$y(0)=0, y'(0)=1$$

I started by doing a reduction of order like so:
$$w=y', w'=y'', \int w=y=\frac{w^{2}}{2}+c$$

which then gave me this:
$$w'=2w(\frac{w^{2}}{2}+c)$$
$$w'=w^{3}+2wc$$

Now I'm stuck on where to go from here. I can't use any of the usual techniques like variation of parameters. There is a trick in my notes here for dealing with equations missing the dependent variable, however I don't fully grasp how it works.

Here is the trick that's given:
For $$yy"=y'^{2}$$
It starts by making z=y'.
Then since you need $$\frac{dz}{dy}$$, you can do this:
$$\frac{dz}{dx}=\frac{dx}{dy}\frac{dy}{dx}$$
then since $$y"=\frac{dz}{dx}$$
you get this: $$y\frac{dz}{dy}=z$$

Does anyone recognize this trick? If so can I apply it here?

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## Answers and Replies

Dick
Science Advisor
Homework Helper
(y^2)'=2yy', right? So your ODE is y''=(y^2)'. Integrate both sides.

I managed to solve it. Thanks for the tip.