Solving for the function $x(t)$ in the differential equation $\frac{dx}{x}=\frac{3}{2}C\left[\frac{\sqrt{A}}{x}+\sqrt{B}\right]dt$ I'm a little stumped on something I'm working on. I have an expression in this following form:

$$\frac{dx}{x}=\frac{3}{2}C\left[\frac{\sqrt{A}}{x}+\sqrt{B}\right]dt$$

I was essentially wondering how I go about solving this for $x$ explicitly.
I'm perhaps thinking I might need to make a substitution which may involve a $\sinh$ function?
Help! 
 A: This equation is separable: By dividing both sides by the quantity in brackets, we get
$$\frac{dx}{\sqrt{A} + \sqrt{B} x} = \frac{3}{2}C \,dt.$$
We can integrate the l.h.s. using the substitution $u = \sqrt{A} + \sqrt{B} x$, $du = \sqrt{B} dx$, and it's easy to invert the resulting equation to give $x$ as a function of $t$. (The constant of parameterization leads to a one-parameter family of solutions.) NB that this technique omits a single solution, corresponding to when the divided quantity is identically zero.
A: Starting with the expression
$$\frac{dx}{x}=\frac32 C\left(\frac{\sqrt{A}}{x}+\sqrt{B}\right)\,dt$$
we divide both sides by $\left(\frac{\sqrt{A}}{x}+\sqrt{B}\right)$ and integrate to find
$$\int \frac{1}{x\left(\frac{\sqrt{A}}{x}+\sqrt{B}\right)}\,dx=\frac32 C\int (1)\,dt$$
Thus, we obtain
$$\frac{1}{\sqrt B}\,\log\left(\sqrt A+\sqrt B x\right)=\frac32 C\,t+K$$
whereupon solving for $t$ yields
$$t=\frac2{3C\sqrt B}\log\left(\sqrt A+\sqrt B x\right)+K'$$
A: $\dfrac{dx}{dt} = Dx + E$ with $D,E$ can be determined from the equation, and you have a linear $1$st order ODE which you have learned and can solve.
