<s>[INST] The differential equation$ \frac{d^2 x}{dt^2}=f(x)$ We are supposed to solve the diff. eq. 
$$x''=\frac{1}{x^3}$$
for $x(0)=1$ and $x'(0)=1$ using a particular approach. I've gotten so far:
Using the interval $I=(0,\infty)$ and $a \in I$ we define $U:I\to \mathbb{R}$:
$$U(x):=-\int_a^x\frac{1}{t^3}dt = \frac{1}{2x^2}-\frac{1}{2a^2}$$
Now we may write 
$$x''=-\frac{dU}{dx}(x)$$
with some simple observations we may see that the function $E$ is constant: 
$$E(t):=\frac{1}{2}(x'(t)^2)+U(x(t))=\frac{1}{2}(x'(0)^2)+U(x(0))=1-\frac{1}{2a^2}$$
So now we may write for $x'$:
$$x'(t)=\sqrt{2(E-U(x(t)))}=\sqrt{2-\frac{1}{x^2}}$$
which is supposibly a differential equation of seperated variables. I don't see the structure of $y'=f(x)g(y)$! How can we get from here to a solution of the original differential equation?
 [/INST] $$\frac{d^2 x}{dt^2}=\frac{1}{x^3}$$
Multiply by $2 \frac{dx}{dt}$ on both sides to get
$$2 \frac{dx}{dt} \frac{d^2x}{dt^2}=2 \frac{dx}{dt} \frac{1}{x^3} \implies \frac{d}{dt} \left( \frac{dx}{dt}\right)^2=\frac{2}{x^3}\frac{dx}{dt}.$$
Next, integrate w.r.t.t both sides
$$\int \frac{d}{dt} \left( \frac{dx}{dt}\right)^2 dt =\int \frac{2}{x^3}\frac{dx}{dt}dt
\implies \left(\frac{dx}{dt} \right)^2 =-\frac{1}{x^2} +A$$
Using $x(0)=1, x'(0)=1$ we get $A=2$, then
$$\frac{dx}{dt}=+\sqrt{2-\frac{1}{x^2}}\ \implies  \int \frac{dx}{\sqrt{2-\frac{1}{x^2}}} =\int t+B$$
$$\implies \int \frac{x}{\sqrt{2x^2-1}}dx=t+B \implies $$
$$\sqrt{2x^2-1}=2t+2B \implies 2B=1$$
Finally we get a hyperbolic trajectory as
$$2x^2-(2t+1)^2=1$$
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