2nd order homogeneous ODE I am trying to solve a system of differential equations (for the full system see below) and I am stuck the following 2nd order ODE (with $a$ and $b$ being constants and $\dot x = \frac{dx}{dt}$):
$$\ddot x - \frac34 a\dot x^2 -b \dot x + 2 a b = 0$$
I tried to substitute $v := \dot x$, which leads me to a different ODE which I was not able to solve. And now I have no idea what else I could try here and would love to get some pointers.
And here the original system / IVP I want to solve:
$$ \dot x = 2 a + 2 e^{ax}y \\ \dot y = \frac32 e^{-ax}a^3+(a^2+b)y-\frac12 a e^{ax}y^2 \\ x(0)=0 \\ y(T)=0$$
In which I solved the first equation for $y$ and substituted $y$ and $\dot y$ in the 2nd which gave me the equation above.

As I said in a comment, start defining $z=x'$; so the differential equation becomes $$\frac{dz}{dt} - \frac34 a z^2 -b z + 2 a b = 0$$ that is to say $$\frac{dz}{dt} = \frac34 a z^2 +b z - 2 a b $$ then, as Semsem suggested, it is separable; so $$\frac{dt}{dz} = \frac{1}{\frac34 a z^2 +b z - 2 a b}$$ so $$t+C=-\frac{2 \tanh ^{-1}\left(\frac{3 a z+2 b}{2 \sqrt{b} \sqrt{6
   a^2+b}}\right)}{\sqrt{b} \sqrt{6 a^2+b}}$$ from which $$z=-\frac{2 \left(\sqrt{b} \sqrt{6 a^2+b} \tanh \left(\frac{1}{2} \sqrt{b} \sqrt{6
   a^2+b} (C+t)\right)+b\right)}{3 a}$$ Now, one more unpleasant integration leads to $$x=-\frac{2 \left(2 \log \left(\cosh \left(\frac{1}{2} \sqrt{b} \sqrt{6 a^2+b}
   (C+t)\right)\right)+b t\right)}{3 a}+D$$
