I tried to solve the following ODE using infinite summs, but failed:
$$ f''(x)+ae^{bx}f(x)=0 \tag1$$
From a comment in this post, I found out that the solution of the above equation contains a linear combination of the Bessel functions. That means that there is a good probability that this equation is solved in detail in some literature.
My question is, does equation $(1)$ have a specific name? I want to google it so I can go through the process of solving it. If not, is there any other way I can find some literature that goes through this problem?
Consider the differential equation: $$ \frac{d^2 y}{d x^2} + a^2 \, e^{b x} \, y = 0. $$ First note that if $a = 0$ then $y(x) = c_{0} + c_{1} \, x$. Second note that if $b = 0$ then $f(x) = c_{0} \, \cos(a x) + c_{1} \, \sin(a x)$. Now consider the case when $a \neq 0$ and $b \ne 0$, which is as follows.
Let $t = e^{b x/2}$ to obtain $\frac{d t}{d x} = \frac{b}{2} \, t$ which is required in \begin{align} \frac{d}{dx} &= \frac{d t}{d x} \, \frac{d}{dt} = \frac{b}{2} \, t \, \frac{d}{d t} \\ \frac{d^2}{d x^2} &= \frac{b^2}{4} \, t \, \frac{d}{d t} \left( t \, \frac{d}{dt} \right) = \frac{b^2}{4} \, \left( t^2 \, \frac{d^2}{d t^2} + t \, \frac{d}{dt} \right) \end{align} and leads to $$ t^2 \, f^{''} + t \, f^{'} + \frac{4 a^2}{b^2} \, t^2 \, f = 0. $$ In order to remove the parameter in the last term let $t = p \, u$ and repeat the change of variables process to obtain, which is $$ \frac{d}{dt} = \frac{1}{p} \, \frac{d}{du} \hspace{6mm} \text{and} \hspace{6mm} \frac{d^2}{dt^2} = \frac{1}{p^2} \, \frac{d^2}{du^2}, $$ and gives $$ u^2 \, f^{''} + u \, f^{'} + \frac{4 \, a^2 \, p^2}{b^2} \, f = 0. $$ Let $$ p^2 = \frac{b^2}{4 \, a^2} \hspace{5mm} \text{which gives} \quad p = \pm \frac{b}{2 a}$$ and the equation $ u^2 \, f^{''} + u \, f^{'} + u^2 \, f = 0$. This equation is the Bessel differential equation of zeroth order and provides the solution as $$ f(u) = A \, J_{0}(u) + B \, Y_{0}(u).$$ Reverting one step yields $$ f(t) = A \, J_{0}\left( \pm \frac{2 a}{b} \, t \right) + B \, Y_{0}\left( \pm \frac{2 a}{b} \, t \right). $$ Note that since $J_{n}(-x)$ is related to $J_{n}(x)$ for $n$ being an integer then choosing the $+$ sign leads to no other solutions to determine. Now reverting the last step then $$ f(x) = A \, J_{0}\left( \frac{2 a}{b} \, e^{b x/2} \right) + B \, Y_{0}\left( \frac{2 a}{b} \, e^{b x/2} \right). $$
From this, the solutions to $y^{''} + a^2 \, e^{b x} \, y = 0$ are $$ y(x) = \begin{cases} c_{0} + c_{1} \, x & a = 0 \\ c_{0} \, \cos(a x) + c_{1} \, \sin(a x) & b = 0 \\ A \, J_{0}\left( \frac{2 a}{b} \, e^{b x/2} \right) + B \, Y_{0}\left( \frac{2 a}{b} \, e^{b x/2} \right) & a \neq 0 \, \text{and} \, b \neq 0 \end{cases}. $$