Solving a first order differential equation in terms of Lambert W-function

Question

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.

Answer 1

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.

Answer 2

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 $$x^2=\frac{c }{b}W\left(\frac{b }{c}e^{\frac{2 a (K+t)}{c}}\right)$$

Answer 3

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.