I came across this question that asked to express solutions to:
$$x^2y'' + xy' + (4x^2 - v^2)y = 0,\quad 0\le x<\infty$$
in terms of Bessel functions subject to the boundary conditions y(x) is bounded in the interval
$0\le x< \infty$ and $y(0) = 5$
The general solution to the Bessel function is $AJ_v(x) + BY_v(x)$ so is that just the answer? I don't know where y(0) = 5 comes into play though...
The general solution of the equation \begin{align} x^{2} y^{''} + x y^{'} + (a^{2} x^{2} - \nu^{2})y = 0 \end{align} is \begin{align} y(x) = A J_{\nu}(ax) + B Y_{\nu}(ax). \end{align} Now, if $y(0) << \infty$ then $Y_{\nu}(ax)$ is not a valid solution because \begin{align} \lim_{x \rightarrow 0} Y_{\nu}(ax) \rightarrow -\infty . \end{align} That is to say that given the boundary condition $y(0) = b$ then \begin{align} y(0) = b = A J_{\nu}(0). \end{align} Solving for $A$ leads to the solution, in this particular case, \begin{align} y(x) = b \ \frac{J_{\nu}(ax)}{J_{\nu}(0)}. \end{align} Notice that this solution is only valid if $\nu = 0$ since $J_{\mu}(0) = 0$ for $\mu > 0$.
