I have an easy-looking task, but I have never solved nonlinear differential equations, so I have no idea how to approach this.

Solve $x'' = x^2$ with inital conditions $x(0) = \sqrt[3]{3}$, $x'(0) = \sqrt{2}$.

I've tried to do a substitution $\frac{\partial x}{\partial t} = y$, then $\frac{\partial^2 x}{\partial t^2} = \frac{\partial y}{\partial x}y$, so we get $y'y = x^2$ and with inital condition $y = \sqrt{1 - x^3}$, But then, when I try to get $x(t)$ back, I finish with integral I cannot compute. Is this approach good? How to finish it?

just hint $$x''x'=x^2x'$$

$$1/2x'^2=1/3x^3+c $$

For time-independent equations, a neat trick is to substitute $v = x'$, then $$x'' = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v \frac{dv}{dx}$$

so you have $$ v\frac{dv}{dx} = x^2 $$

Using separation of variables we obtain $$ v^2 = \frac{2x^3}{3} + C $$ We have $x = \sqrt[3]{3} \implies v = \sqrt{2}$, so $C=0$, and $v$ must take the positive root $$ v(x) = \frac{dx}{dt} = \sqrt{\frac{2}{3}}x^{3/2} $$

You can use separation of variables again to get the final solution, which is $$ x(t) = \left( 3^{-1/6}-\sqrt{\frac{3}{2}}t \right)^{-2} $$

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