How do I find an analytical solution of $x x' + \frac 3 2 (x')^2 + A \frac {x'} x = W$? I am trying to solve the Rayleigh-Plesset equation of a bubble made of super-critical fluid, and its radius varies as a function of time. A generalised form of the equation I got is:
$$x x' + \frac 3 2 (x')^2 + A \frac {x'} x = W .$$
Here $A$ and $W$ are constants. How do I go about solving this system?

$$x x' + \frac 3 2 (x')^2 + A \frac {x'} x = W .$$
We observe that $x(t)$ is an even function. This draw to the change of function :
$$X=x^2 \quad\to\quad dX=2xdx \quad\to\quad x'=\frac{dx}{dt}=\frac{dx}{dX}\frac{dX}{dt}=\frac{1}{2x}\frac{dX}{dt}$$
$$x \left(\frac{1}{2x}\frac{dX}{dt} \right) + \frac 3 2 \left(\frac{1}{2x}\frac{dX}{dt} \right)^2 + \frac {A}{x} \left(\frac{1}{2x}\frac{dX}{dt} \right)= W .$$
$$ 3 \left(\frac{dX}{dt} \right)^2 + 4(X+A)\frac{dX}{dt} - 8WX =0 $$
$$\frac{dX}{dt}=\frac{2}{3}\left(X+A\pm\sqrt{(X+A)^2+6WX } \right)$$ 
$$t=\frac{3}{2}\int \frac{dX}{X+A\pm\sqrt{(X+A)^2+6WX } }$$
$$t=t_0+\frac{3}{2}\int_{x_0^2}^{x^2} \frac{dX}{X+A\pm\sqrt{(X+A)^2+6WX } }$$
This will give $t(x)$ on form of an huge formula. It's doubtful that the inverse function $x(t)$ could be expressed on closed form.
