Solving this equation : $ y'' + (1/x)y' - (k^2 - (m^2/x^2) )y=0$ gives $y(x) = J_m (-ikx) + Y_m (-ikx)$ which is further written as $y(x) = \frac{1}{\sqrt{x}}\left(\exp(-kx) + \exp(kx) \right)$
Here $J_m, Y_m$ are Bessel functions of first and second kind respectively & $i = \sqrt{-1}$.
Please explain how Bessel functions are converted to exponential form.
Thanks in advance.
Some notes:
The differential equation \begin{align} \left[ x^{2} D_{x}^{2} + x D_{x} - ((kx)^{2} - m^{2}) \right] y =0 \end{align} has the solution \begin{align} y(x) = A J_{im}(-ikx) + B Y_{im}(-i kx) \end{align} where $J_{n}(x)$ and $Y_{n}(x)$ are Bessel functions. Notice that these functions are of order $i m$ which are complex in value.
For the case of $m$ being of fractional order the differential equation \begin{align} \left[ x^{2} D_{x}^{2} + 2 x D_{x} + (x^{2} - n(n+1)) \right] y = 0 \end{align} has the solution \begin{align} y(x) = A j_{n}(x) + B \mathcal{y}_{n}(x). \end{align}
Further information on the Bessel functions can be obtained, often the starting point these days, from Bessel Functions (Wiki).
As given by the question the solution is \begin{align} y(x) = \frac{2 \, \cosh(kx)}{\sqrt{x}}. \end{align} This leads to the differential equation \begin{align} x^{2} y'' + x y' - ((kx)^{2} + \frac{1}{4})y = o \end{align} which is a modified Bessel function of fractional order and has the form \begin{align} y(x) = A I_{-1/2}(kx) + B I_{1/2}(kx) = A \sqrt{\frac{2}{k\pi x}} \cosh(kx) + B \sqrt{\frac{2}{k\pi x}} \sinh(kx). \end{align} From this it is seen that $B=0$ and $A = \sqrt{2 k \pi}$.
Another example of similarity:
Consider the solution \begin{align} y(x) = \frac{2}{x^{5}} \, \cosh(kx) \end{align} satisfies the differential equation \begin{align} x^{2} y'' + 10 x y' -((kx)^{2} - 20)y = 0. \end{align} Making the substitution \begin{align} y(x) = \frac{1}{x^{4}} \, g(x) \end{align} it is seen that $g(x)$ satisfies $x^{2} g'' + 2 x g' -(kx)^{2} g = 0$ which has the solution \begin{align} g(x) = A j_{0}(-ikx) + B \mathcal{y}_{0}(-ikx) = \frac{1}{x} (A_{1} \sinh(kx) + B_{1} \cosh(kx) ). \end{align} This leads to \begin{align} y(x) = \frac{1}{x^{5}} \left[ A_{1} \sinh(kx) + B_{1} \cosh(kx) \right]. \end{align} Now, letting $A_{1} = 0$ and $B_{1} = 2$ it is then seen that \begin{align} y(x) = \frac{C}{x^{4}} \, \mathcal{y}_{0}(-ikx). \end{align}