$x''= \frac{Ax+B}{Cx+D}$

Question

Might there be a closed-form solution to the second-order differential equation below?$$x''(t)=\frac{Ax+B}{Cx+D}$$ If not, is there any way to get a power series approximation in terms of the variables, without having to use any initial conditions? Or absolutely anything somewhat close to the answer? (If it helps we have $x(0)=x'(0)=0$ but the constants $A,B,C,D$ are unknown and must be left as variables.)

Answer 1

Let $v=x'(t)$ and $a=x''(t)$:

Then $$a=v\frac{dv}{dx}=\frac{Ax+B}{Cx+D}\\ \int vdv=\int \frac{Ax+B}{Cx+D} dx\\ \frac{v^2}2=\frac{(BC-AD)\ln(Cx+D)+ACx}{C^2}+k\\ v=\frac{dx}{dt}=\sqrt{2\frac{(BC-AD)\ln(Cx+D)+ACx}{C^2}+k'}\\ \int \frac{dx}{\sqrt{2\frac{(BC-AD)\ln(Cx+D)+ACx}{C^2}+k'}}=\int dt\\f(x)=t+c$$ I suppose there is no closed form of $f$