Solving a first order differential equation in terms of Lambert W-function I am having great difficulty solving the following equation
$$\  \frac{ax}{(bx^2 + c)} = \frac{dx}{dt}  $$
I have re-edited the question. Any help is appreciated. Thank you.
 A: This was for the first edit of the post.
Welcome to the world of Lambert function !
The solution of equation $$\ x^2 + \log(x) - c = 0$$ is given by 
$$x=\pm\frac{\sqrt{W\left(2 e^{2 c}\right)}}{\sqrt{2}}$$ but only the positive root must be kept because of $\log(x).
The Wikipedia page gives series espansions fot he evaluation of $W(.)$.
If you cannot use Lambert function, think that you are looking for the zero of function $$f(x)=x^2 + \log(x) - c$$ $$f'(x)=2x+\frac 1x$$  The first derivative does not cancel in the real domain and it is always positive; so, only one root that you could find numerically using any method (secant, Newton, ...). For starting,  you could start iterating using $x_0=\sqrt c$.
Suppose $c=0.01$. Newton iterations would then  be
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 0.1000000000 \\
 1 & 0.3257436366 \\
 2 & 0.6013225902 \\
 3 & 0.6561217076 \\
 4 & 0.6564436996 \\
 5 & 0.6564437054
\end{array}
\right)$$
For the new version of the post.
Considering the differential equation
$$  \frac{ax}{(bx^2 + c)} = \frac{dx}{dt}\implies \frac{dt}{dx}=\frac{b x}{a}+\frac{c}{a x}$$ we then have $$t+K=\frac{b x^2}{2 a}+\frac{c \log (x)}{a}$$ which would then lead to 
$$x^2=\frac{c }{b}W\left(\frac{b }{c}e^{\frac{2 a (K+t)}{c}}\right)$$
A: For the equation
$$\frac{ax}{(bx^2 + c)} = \frac{dx}{dt}$$
it can be seen that
\begin{align}
\frac{dx}{dt} &= \frac{a}{2 b} \, \frac{2 b x}{b x^{2} + c} = \frac{a}{2 b} \, \frac{d}{dt} \, \ln(b x^{2} + c )
\end{align}
which yields 
$$x(t) = \frac{a}{2 b} \, \ln(b x^{2} + c) + c_{0}.$$
Solving for $x(t)$ requires much struggle to obtain
$$x^{2}(t) = \frac{c}{b} \, W\left(\frac{b}{c} \, e^{\frac{2 \, a}{c} \, (t + p_{0})} \right),$$
where $p_{0}$ is a constant, and $W(x)$ is the Lambert w-function. 
For verification use:
$$\frac{dW}{d z} = \frac{W(z)}{z \, (1 + W(z))}$$
for which
\begin{align}
2 \, x \, \frac{dx}{dt} &= \frac{c}{b} \, \frac{b}{c} \, \frac{2 a}{c} \, e^{\frac{2 \, a}{c} \, (t + p_{0})} \, W'(\cdot) \\
&= \frac{2 a}{c} \, e^{\frac{2 \, a}{c} \, (t + p_{0})} \, \frac{W(\cdot)}{\frac{b}{c} \, e^{\frac{2 \, a}{c} \, (t + p_{0})} \, (1 + W(\cdot))} \\
&= \frac{2 a}{b} \, \frac{W(\cdot)}{1 + W(\cdot)} \\
&= \frac{2 a}{b} \, \frac{\frac{b}{c} \, x^{2}}{1 + \frac{b}{c} \, x^{2}} \\
&= \frac{2 a \, x^{2}}{b \, x^2 + c} \\
\frac{dx}{dt} &= \frac{a \, x}{b \, x^{2} + c}.
\end{align}
This is the first order differential equation that was given.
A: Please note that $$\ln(x)=y \implies x=e^y$$
So I think what you are asking is solved using rearrangement and then using the above property. There is no differential equation here. 
