Arnol'd's trivium problem 21 Find the derivative of the solution of the equation $\ddot{x} = \dot{x}^2 + x^3$ with initial condition $x(0)=0$, $\dot{x}(0) = A$ with respect to A for A = 0.
I have no idea about how to solve this problem, please help me.
 A: The general procedure is:


*

*solve the equation with the given the initial conditions. Here the conditions are $x(0)=0$, $\dot x(0)=0$, the solution is $x_0(t)=0$.

*linearize the equation around $x_0$. That is: plug $x(t)=x_0(t)+\epsilon y(t)$ to the differential equation, and keep only the part linear in $\epsilon$ (i.e. we compute a derivative). Here we get $\ddot y=0$, with the general solution $y(t)=a+bt$.
Now to compute the derivative of $x(t)$ by $A$ we just need to take the $y$ s.t. $y(0)=0$ and $\dot y(0)=1$; in this case it is 
$$y(t)=t.$$ 
A: Let $ w(x) = \dot{x}^2 $, and denote differentiation of $w$ w.r.t. $x$ by $w'$.
$ w'(x) = 2\dot{x} \frac{d \dot{x}}{dx} = 2\dot{x} \frac{d \dot{x}}{dt}(\frac{dx}{dt})^{-1} = 2\dot{x}\space\ddot{x}\space\dot{x}^{-1} = 2\ddot{x} $
With this substitution the ODE in question reduces to the following linear first order equation: 
$$ w' - 2w = 2x^3 .$$
Solving for $w$ as a function of $x$ will then give you the autonomous first order equation $ \dot{x}^2 = w(x). $
