Analytical solution to equation $ \arctan(x)-k \arctan(x/k)=c$

Question

For the equation:$$\arctan(x)-k \arctan(x/k)=c$$ which is part of a gasdynamics function called Prandtl–Meyer function, it is not difficult to find the solution numerically, however, I'm wondering, does the analytical solution exists? Thanks.
An example for the coefficient is $k=2.4$, $c=-1.4$, the solution is $x=5.57$.

Answer 1

After substituting $a\to i a,b\to ib,c\to e^{-2i c}$ and simplifying the Lagrange inversion:

$$e^{2a\tanh^{-1}(bx)-2\tanh^{-1}(x)}=c\implies x=1+\sum_{n=1}^\infty \frac{(-c)^n}{n!}\left.\frac{d^{n-1}}{dw^{n-1}}\left(\left((w+1)(1-bw)^a (bw+1)^{-a}\right)^n\right)\right|_{w=1}$$

shown here. Now to use the generalized product rule and expanding. The result will be a double hypergeometric series converging for values similar to the “shown here” link. More will develop in the series expansion.

Answer 2

The fuction $$f(x)=k \tan ^{-1}\left(\frac{x}{k}\right)-\tan ^{-1}(x)-c$$ being odd, just consider the positive roots for $k>1$ and $c>0$.

The function being rather flat around the origin, if $c$ is "sufficiently" large, using Taylor series and power series reversion to obtain $$x=t-\sum_{n=0}^p \frac{P_n(k)}{b_n}\,\frac 1{t^{2n+1}} \quad \text{where}\quad t=\frac{2(k^2-1)}{(k-1)\pi-2c} $$

Coefficients $b_n$ correspond to the $a_{n+1}$ in sequence $A036278$ at $OEIS$ and the first polynomials are $$\left( \begin{array}{cc} n & P_n(k) \\ 0 & (k^2+1) \\ 1 & k^4+11 k^2+1 \\ 2 & 2 \left(k^2+1\right) \left(k^4+56 k^2+1\right) \\ 3 & k^8+247 k^6+723 k^4+247 k^2+1 \\ 4 & 2 \left(k^2+1\right) \left(k^8+1012 k^6+5028 k^4+1012 k^2+1\right) \\ \end{array} \right)$$ We can easily as many as desired since the coefficients of the intial series are more than simple

$$k \tan ^{-1}\left(\frac{x}{k}\right)-\tan ^{-1}(x)=\sum_{n=0}^\infty (-1)^n \frac{k^{2(n+1)}-1}{(2n+1)\, x^{2n+1}}$$