Find $x$ such that $\cosh(a + bx) + 1 = cx$

Question

I need to find an analytical solution for $x$ to: $$ \cosh(a + bx) + 1 = cx $$

where a,b and c are real parameters.

I have tried to tackle this geometrically, by splitting the problem into finding the intersection of two curves, one being $y=\cosh(a + bx) + 1$ and the other being $y = cx$, but with no luck.

I tried to convert to polar form but I cannot seem to convert $y = \cosh(a+bx)$ easily.

The solution I am looking for must be in the form $x = \dots$. Ideally $x$ should be real but I'm not too concerned if it turns out to be complex.

Thanks for any help provided.

Answer 1

A suggestion: take $u=a+bx \ $ as a substitution
$$x=\frac{u-a}{b}\to \\ cosh(a+bx)+1=cx\\cosh(u)+1=c\frac{u-a}{b}\\e^u+e^{-u}+2=2c\frac{u-a}{b}$$

Answer 2

I tried an approach using the Laplace transform which goes as follows. $cosh(a+bx):=\frac{e^{(a+bx)}+e^{-(a+bx)}}{2}=cx-1$. Now $e^{(a+bx)}+e^{-(a+bx)}=2cx-2$. Then taking the Laplace transform on both sides we get: $e^{a}×\frac{1}{s-b}+e^{-a}×\frac{1}{s+b}=2c[\frac{1}{s^{2}}]-2[\frac{1}{s}]$. We can rewrite: $\frac{e^{a}}{(s-b)}+\frac{e^{-a}}{(s+b)}=\frac{2c}{s^{2}}-\frac{2}{s}$. Now we simply do common denominators as follows: $\frac{e^{a}(s+b)+e^{-a}(s-b)}{(s^{2}-b^{2})}=\frac{2c-2s}{s^{2}}=2(\frac{c-s}{s^{2}})$. I don't know where it may lead, or if it is useful in itself, but I wanted to share it because someone can maybe take the idea and solve the problem using the Laplace transform.

Answer 3

Substitute $ax+b=w$ to simplify the equation:

$\cosh(ax+b)+1=cx\iff \cosh(w)=\frac cbw-\frac{ac}b-1=uw+v$

We use Lagrange reversion to get a real root:

$$\cosh(w)-uw=v\implies w=-\frac vu+\sum_{n=1}^\infty\frac1{u^nn!}\left.\frac{d^{n-1}}{dw^{n-1}}\cosh^n(w)\right|_{-\frac vu}$$

Finally, applying binomial theorem or a Wolfram functions identity gives:

$$\cosh(w)-uw=v\implies w=-\frac vu+\sum_{n=1}^\infty\sum_{m=0}^n\frac{(2m-n)^{n-1}e^{\frac vu(n-2m)}}{(2u)^n(n-m)!m!}$$

shown here