Find the particular solution from this Modified Bessel D.E \begin{align} x^2\frac{d^2u}{dx^2} + x\frac{du}{dx} - \alpha^2x^2u(x) = 0, \hspace{10pt}x\in[a,b] \end{align} with boundary condition $u(a)=1,u'(b)=0$
My attempt: Let $y=\alpha x$, then $u(x)=v(y)$, and the D.E becomes \begin{align} y^2\frac{d^2v}{dy^2} + y\frac{dv}{dy} -y^2 v(y) = 0 \end{align} So the general solution is \begin{align} v(y) & = A I_0(y) + BK_0(y) \\ \Rightarrow u(x) & = AI_0(\alpha x)+BK_0(\alpha x) \end{align} where $I_0(y)$ and $K_0(y)$ represent zero order of Modified Bessel function first and second kind. Now my problem is to find the particular solution \begin{align} u(a) & = A I_0(\alpha a) + BK_0(\alpha a)= 1 \\ u'(b) & = A\alpha I'_0(\alpha b) + B\alpha K'_0(\alpha b)=0\end{align} How can I find $A$ and $B$?
Edit:
Since $I_0(\alpha a), K_0(\alpha a), I'_0(\alpha b), K'_0(\alpha b)$ is a constant, the equations above can be treated as a system of linear equation of two variables.
First of all, the solution is not short at all. It is actually quite large so let me give you the main ideas of the solution. First of all, start by writing the solution as a power series $$ \qquad \qquad u=\sum_{n=0}^{+\infty} a_nx^n, \qquad \qquad (*) $$ where the coefficients $\{a_n\}_{n\in\mathbb{N}}$ have to be determine! To determine the coefficients we will use the equation, but first, let's do some preliminary computations: First, note that by directly taking the derivative into $(*)$ we have $$ xu'=x\sum_{n=0}^{+\infty} na_nx^{n-1}=\sum_{n=0}^{+\infty}na_{n}x^n $$ and in the same way we also have $$ x^2u''=x^2\sum_{n=0}^{+\infty}n(n-1)a_nx^{n-2}=\sum_{n=0}^{+\infty}n(n-1)a_{n}x^{n}. $$ Note that the first sum actually starts at $n=1$ and by the same reasons the latter one starts at $n=2$ (because the coefficients associated to $n=0$ and $n=0,1$ respect. become equals zero). Finally, notice that $$ x^2u=\sum_{n=0}^{+\infty}a_nx^{n+2}=\sum_{n=2}^{+\infty}a_{n-2}x^n $$ Now, replacing all of these formulas into the equation we obtain that if $u$ is a solution then the following holds: $$ a_1x+\sum_{n=2}^{+\infty }\big(n(n-1)a_n+na_n-\alpha^2a_{n-2}\big)x^n\equiv0 $$ So we obtain an identically zero polynomial. The only way that this can happen is that each of the coefficients are zero simultaneously. In other words, we have the system $$ \begin{cases} a_1=0, \\ n(n-1)a_n+na_n-\alpha^2a_{n-2}=0 & \hbox{for all }\, n=2,3,.... \end{cases} $$ Then, you need to solve this system. In order to do this, you have to solve the recursion for $a_n$ in the latter identity (that is, write $a_n$ in terms of $a_{n-2}$ and then replace the $a_{n-2}$ by its formula containing $a_{n-4}$ and so forth). Now, since this is becoming quite long I let you the final details. This shouldn't be that difficult.
Edit: Of course notice that as a final step you will have to use your initial conditions to obtain the explicit values of the coefficients $a_n$ (just after you solve the recursion). This will give you the explicit solution of the equation. Luckily, you will be able to recognize the series of an well-known function (for instance the series of the $\mathrm{sech}(x)$ or other function).