# Second Order differential equation problem

I this trying to solve $y'' = yy'$ with $y(0) = 1, y'(0) = 1.$

I know $y' = 1/x'$, $y'' = -x''/(x')^3$ . Then I tried substitute $u = x'$. So $du = x''$. $y'' = -du/u^3$. Then $\int y'' = \frac{1}{2u^2}$. I am a bit confused at this point. Could someone point out how to proceed?

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One way is to integrate both sides. Then things should look familiar. – André Nicolas Oct 13 '11 at 5:53
@AndréNicolas: Are you talking about $y'' = yy'$? – Mark Oct 13 '11 at 5:54
Yes, the original equation. – André Nicolas Oct 13 '11 at 5:58
@André is referring to the fact that both $y''$ and $yy'$ are the derivatives of simple functions of $y'$ and $y$. – Did Oct 13 '11 at 5:58
What does "$y'=1/x'$" mean? Is that your way of saying $dy/dx=1/(dx/dy)$? – Gerry Myerson Oct 13 '11 at 6:07

Even though this can be done just by integrating both sides, the following technique might be of interest.. sometimes for such equations it helps to write $y'(x) = z(y(x))$. Taking derivatives you get $y''(x) = z'(y(x)) y'(x) = z'(y) z$. So the equation $y'' = y y'$ becomes, when viewed as functions of $y$: $$z'(y) z = yz$$ Which is the same as $$z'(y) = y$$ So $z(y) = {1 \over 2}y^2 + C$ So you are reduced to solving $${dy \over dx} = {1 \over 2}y^2 + C$$ You can separate variables to solve this.