I encounter the following second-order ODE, but not sure how to approach it. Can somebody help?
$y''(x) = f(x) y(x)$ for $x>0$ where $f(x)=\frac{1}{2x^{3/2}} + \frac{1}{x} + C$.
Solve $y(x)$ for $x>0$. $y(x)$ is symmetric, that is, $y(-x)=y(x)$. $C$ is the eigen value such that $y(x)$ is finite at infinity.
Thanks.
$y(x)'' = f(x) y(x)\quad$ where $\quad f(x)=\frac{1}{2x^{3/2}} + \frac{1}{x} + C$.
I doubt that a closed form exists for $y(x)$.
Some closed forms can be found in simplified cases :
$f(x)=\frac{1}{x} \quad\to\quad y(x)$ involves Bessel functions.
$f(x)=\frac{1}{x}+C \quad\to\quad y(x)$ involves Confluent Hypergeometric functions.
$f(x)=\frac{1}{2x^{3/2}} \quad\to\quad y(x)$ involves Bessel functions.
In case of $\quad f(x)=\frac{1}{2x^{3/2} } +\frac{1}{x} \quad$ it seems that no closed form exist involving standard special functions.