Proving the existence of an $x$ such that $(x-a)(x-b)e^{cx} = (x+a)(x+b)e^{-cx}$

Question

I've been working on a PDE problem and I've brought it to the following form: $$ (x-a)(x-b)e^{cx} = (x+a)(x+b)e^{-cx} $$ Where $a, b, c,$ and $x$ are all $> 0$. I'm looking to prove that for any $a, b, c$ fixed we can find an $x$.

I have no idea how to start with this one! Any advice? I tried messing around with the first and second derivatives, but ended up not getting anywhere useful.

Thanks!

Answer 1

Hint Let $f(x)=(x-a)(x-b)e^{cx} - (x+a)(x+b)e^{-cx}$.

Then $f(a)<0$ (or $f(b) <0$) and $\lim_{x \to \infty} f(x) = \infty$.

Use the Intermediate value theorem.