I have had some excellent answers in response to a previous question here:
Solving a first order differential equation in terms of Lambert W-function
However I would like to show my work some more since I am still stuck, and terribly lost, in hopes for some further guidance. Thank you all.
So I was left with an equation of the following form:
$$ t + K = \frac{bx^2}{2a} + \frac{c\ln(x)}{a} $$
Multiplying through by $$ \frac{2a}{c}$$ Gives
$$ \frac{2a(t + K)}{c} = \frac{b}{c} x^2 + 2\ln(x) $$
Following by:
$$ exp(\frac{2a(t + K)}{c}) = x^2 exp(bx^2/c) $$
It is at this point here I am stuck. I actually am not 100 percent sure the final step I made is valid. However I belive here that the function is now in the form: $$ y = x exp(x) $$ Which is the form needed to use the Lambert function $$ W(y) = x $$
Any help is appreciated. Thanks all.
From $$exp(\frac{2a(t + K)}{c}) = x^2 exp(bx^2/c)$$ multiply by $b/c$ and let $y = (b/c) \, x^{2}$ for $$\frac{b}{c} \, exp(\frac{2a(t + K)}{c}) = y exp(y).$$ Now, use the Lambert W-function to obtain, $y = z \, e^{z} \to z = W(y)$, $$y = W\left(\frac{b}{c} \, exp(\frac{2a(t + K)}{c})\right)$$ and since $y = (b/c) \, x^{2}$ then $$x^{2} = \frac{c}{b} \, W\left(\frac{b}{c} \, exp(\frac{2a(t + K)}{c})\right).$$