inverse a function with exponential and first degree polynom

Question

I need some help to inverse this function: $$ y = a(e^{bx}-1) + cx + d $$ with $y(0)=d$ and $y(k)=0$ where $k$ is a constant.

I don't know how to proceed.

Thanks.

Answer 1

First you should check where the function is actually invertible. By making the derivative: $$ y'(x) = ab e^{bx} + c $$ you can see that the derivative is null in $x_0 = \log(-c/(ab))/b$ so that the function is invertible either in the interval $[x_0, +\infty)$ or in $(-\infty, x_0]$. Once you have decided which is the interval you need you can adopt any algorithm to find the root of a function (the simplest being bisection method) and use it to solve the equation $y(x) = y$.

Answer 2

There is a chance that tis answer be out off topic; so forgive me if this is the case.

Any equation which write or can rewrite $$A+Bx+C\log(D+Ex)=0$$ has solution(s) in terms of Lambert function $W(t)$ (as Michael Galuza commented) which by definition corresponds to $$t=W(t)\,e^{W(t)}$$ In the case of $$ a(e^{bx}-1) + cx + d-y=0$$ the solution is given by $$x=\frac{a-d+y}{c}-\frac{1}{b}W(z)$$ where $$z=\frac{a b }{c}e^{\frac{b (a-d+y)}{c}}$$

But I must confess that I do not see where the conditions $y(0)=d$ and $y(k)=0$ could apply. The first one is always satisfied and the second one could only be used to fix one of the other parameters.