Solution of $\frac{dx}{dt}=-\frac{(\sigma+1)x}{\sigma x+1}$ in terms of Lambert $w$ function

Question

Solution of $\frac{dx}{dt}=-\frac{(\sigma+1)x}{\sigma x+1}$ in terms of Lambert $w$ function.

Should I first take the solution of ODE and then apply Laplace transform. Please give step by step solution if possible

Answer 1

lets take $\sigma x = \phi$ then $$ \dfrac{dx}{dt} = -\dfrac{(\sigma+1)x}{\sigma x + 1} $$ transforms to $$ \dfrac{d\phi}{dt} = -(\sigma+1)\dfrac{\phi}{\phi+1} $$ re-arrange leads to $$ \int 1+\dfrac{1}{\phi}d\phi = \phi + \ln \phi = \int -(\sigma+1) dt = -(\sigma+1)t + C $$ thus $$ \phi\mathrm{e}^{\phi} = C_1\mathrm{e}^{-(\sigma+1)t} $$ when t = 0 we find $$ C_1 = \phi(t=0)\mathrm{e}^{\phi(t=0)} = \sigma x(t=0) \mathrm{e}^{\sigma x(t=0)} $$ thus $$ \phi\mathrm{e}^{\phi} = \sigma x(t=0) \mathrm{e}^{\sigma x(t=0)} \mathrm{e}^{-(\sigma+1)t} $$ thus

$$ \phi\mathrm{e}^{\phi} = \sigma x(t=0) \mathrm{e}^{\left(\sigma x(t=0) -(\sigma+1)t\right)} $$ now using a standard result $$ Y = X\mathrm{e}^X \implies X = W(Y) $$ thus $$ \phi\mathrm{e}^{\phi} = \sigma x(t=0) \mathrm{e}^{\left(\sigma x(t=0) -(\sigma+1)t\right)} $$ leads to $$ \phi = W\left[\sigma x(t=0) \mathrm{e}^{\left(\sigma x(t=0) -(\sigma+1)t\right)}\right] $$ now transforming back to $x$ we find

$$ x(t) = \dfrac{W\left[\sigma x(t=0) \mathrm{e}^{\left(\sigma x(t=0) -(\sigma+1)t\right)}\right]}{\sigma} $$

the finally piece of your puzzle is the condition that $x(t=0)=1$ to get the result that you show. If thats the case then $$ x(t) = \dfrac{W\left[\sigma \mathrm{e}^{\left(\sigma -(\sigma+1)t\right)}\right]}{\sigma} $$