Solving an exponential equation with variable in linear sum

Question

I have the following equation: $$x(1+c-ae^x)=y(1+c-be^y)$$ where $a, b, c$ are constants.

Can I find a closed-form expression of $y$ in terms of $x$ and constants?

Thanks!

Answer 1

Your equation is an exponential-polynomial equation.

$$x(c+1-ae^x)=y(c+1-be^y)$$ $$-axe^x+cx+x=-bye^y+cy+y$$ $$-axe^x+cx+x+bye^y=cy+y$$ $$bye^y=cy+y+axe^x-cx-x$$ $$\frac{by}{cy+y+axe^x-cx-x}e^y=1$$ $$\frac{by}{(c+1)y+axe^x-cx-x}e^y=1$$ $$\frac{y}{\frac{(y+1)y+axe^x-cx-x}{c+1}}e^y=\frac{c+1}{b}$$ $$\frac{y}{y+\frac{axe^x-cx-x}{c+1}}e^y=\frac{c+1}{b}$$ $$\frac{y}{y-\frac{-axe^x+cx+x}{c+1}}e^y=\frac{c+1}{b}$$

We see, your equation cannot be solved in terms of Lambert W in the general case. But it can be solved in terms of Generalized Lambert W:

$$y=W\left(^{\ \ \ \ \ \ \ \ 0}_{\frac{-axe^x+cx+x}{c+1}};\frac{c+1}{b}\right)=W\left(^{\frac{axe^x-cx-x}{c+1}}_{\ \ \ \ \ \ \ \ 0};\frac{b}{c+1}\right)$$

$-$ see the references below.

For $c=-1$, we have the equation $axe^x=bye^y$ with the solutions $y=W\left(\frac{axe^x}{b}\right)\ \ \ (k\in\mathbb{Z})$.

[Mezö 2017] Mezö, I.: On the structure of the solution set of a generalized Euler-Lambert equation. J. Math. Anal. Appl. 455 (2017) (1) 538-553

[Mezö/Baricz 2017] Mezö, I.; Baricz, A.: On the generalization of the Lambert W function. Transact. Amer. Math. Soc. 369 (2017) (11) 7917–7934 (On the generalization of the Lambert W function with applications in theoretical physics. 2015)

[Castle 2018] Castle, P.: Taylor series for generalized Lambert W functions. 2018

Answer 2

Without knowing other information, we can only speculate about, but one thing is certain: you cannot get any closed form solution for $y$ in terms of $x$, not even using Lambert $W$ function.

What I do suggest is a sort of "amusement run", where we take the exponential in $y$ and we approximate it with its Taylor series, increasing the order.

1

We start with $e^y \sim 1 + y + O(y^2)$.

For the sake of brevity, we will call $T(x) = x(1+c-ae^x)$

Then we have to solve

$$T(x) = y(1 + c - b(1+y))$$

we just get a quadratic equation in $y$, whose solution is

$$y = \frac{-b+c+1}{2 b}\pm\frac{1}{2} \sqrt{\frac{b^2-2 b c-4 b T(x)-2 b+c^2+2 c+1}{b^2}}$$

2

We increase the order of the Series: $e^y \sim \ + y + \frac{y^2}{2} + O(y^3)$ and we get a cubic painful equation:

$$T(x) = y(1 + c - b(1 + y + y^2/2))$$

whose solutions are

$$y_0 = \frac{\sqrt[3]{20 b^3-36 b^2 k-54 b^2 T+\sqrt{4 \left(2 b^2-6 b k\right)^3+\left(20 b^3-36 b^2 k-54 b^2 T\right)^2}}}{3 \sqrt[3]{2} b}-\frac{\sqrt[3]{2} \left(2 b^2-6 b k\right)}{3 b \sqrt[3]{20 b^3-36 b^2 k-54 b^2 T+\sqrt{4 \left(2 b^2-6 b k\right)^3+\left(20 b^3-36 b^2 k-54 b^2 T\right)^2}}}-\frac{2}{3}$$

$$y_1 = -\frac{\left(1-i \sqrt{3}\right) \sqrt[3]{20 b^3-36 b^2 k-54 b^2 T+\sqrt{4 \left(2 b^2-6 b k\right)^3+\left(20 b^3-36 b^2 k-54 b^2 T\right)^2}}}{6 \sqrt[3]{2} b}+\frac{\left(1+i \sqrt{3}\right) \left(2 b^2-6 b k\right)}{3\ 2^{2/3} b \sqrt[3]{20 b^3-36 b^2 k-54 b^2 T+\sqrt{4 \left(2 b^2-6 b k\right)^3+\left(20 b^3-36 b^2 k-54 b^2 T\right)^2}}}-\frac{2}{3}$$

$$y_2 = -\frac{\left(1+i \sqrt{3}\right) \sqrt[3]{20 b^3-36 b^2 k-54 b^2 T+\sqrt{4 \left(2 b^2-6 b k\right)^3+\left(20 b^3-36 b^2 k-54 b^2 T\right)^2}}}{6 \sqrt[3]{2} b}+\frac{\left(1-i \sqrt{3}\right) \left(2 b^2-6 b k\right)}{3\ 2^{2/3} b \sqrt[3]{20 b^3-36 b^2 k-54 b^2 T+\sqrt{4 \left(2 b^2-6 b k\right)^3+\left(20 b^3-36 b^2 k-54 b^2 T\right)^2}}}-\frac{2}{3}$$

Where here $k = 1+c$.

As you see it doesn't get easier. I let you imagine what would be the third order...

Also consider that those solutions are not (or could not be) even approximations of the "real" solutions (assuming to deal with $a, b, c$ fixed numbers) for we are treating the exponential first like a linear function, and secondly like a quadratic.

Answer 3

COMMENT.- Just for fun, $x(d-ae^x)=y(d-be^y)\iff \dfrac yx=\dfrac{d-ae^x}{d-be^y}$ so for a real $t$ we have $y=(d-ae^x)t$ and $x=(d-be^y)t$. It follows $$y=\ln\left(\frac{dt-x}{bt}\right)=(d-ae^x)t$$ Consequently for each value $x_0$ we have a function $$f_{x_0}(t)=\frac{\ln(A-\frac Bt)}{C}$$ where $A,B,C$ are non-independent constants. Graphically we can verify that when this function is well defined it cuts the diagonal just one time and this unique fixed point $t_0$ let us to do $$y_{0}=(d-ae^{x_0})t_0$$

Obviously the task is hard but I have say "just for fun" (and this is just a comment!).

Answer 4

In terms of possible approximation, use (assuming that $1+c \neq b$) $$f(x)=y \left((1+c)-b e^y\right)=(1+c-b)y-b\sum_{n=1}^\infty\frac{y^n}{(n-1 )!}$$ and use series reversion to obtain $$y=t+\frac{b }{1+c-b}t^2+\frac{b (1+c+3 b)}{2 (1+c-b)^2}t^3+\frac{b \left(16 b^2+13 b (1+c)+(1+c)^2\right)}{6 (1+c-b)^3}t^4+O(t^5)$$ where $$t=\frac{f(x)}{1+c-b}$$