Answer
$x=-\ln (e^k-e^{2t})+C$
Work Step by Step
We are given:
$\frac{d^2x}{dt^2}=(\frac{dx}{dt})^2+2\frac{dx}{dt}$
Suppose that
$u=\frac{dy}{dx}$
Now, $\frac{du}{dt}=u^2+2u$
$\frac{du}{u^2+2u}=dt$
Integrating both sides:
$\frac{1}{2}\int (\frac{du}{u}-\frac{du}{u+2})=\int dt$
$ \frac{1}{2}\ln \frac{u}{u+2}=t+C$
$\ln \frac{u}{u+2}=e^{2(t+C)}$
$1+\frac{2}{u}=e^{-2(t+C)}$
Hence,
$u=\frac{2}{e^{-2(t+C)}-1}$
$\frac{dx}{dt}=\frac{2}{e^{-2(t+C)}-1}$
$dx=\frac{2}{e^{-2(t+C)}-1}dt$
$dx=\frac{2}{2^{k-2t}-1}dt$
Integrating both sides:
$\int dx =\frac{2}{2^{k-2t}-1}dt $
The final solution is:
$x=-\ln (e^k-e^{2t})+C$
