I was asked the following question on my homework assignment:
We have the ODE $x''(t) + \frac{2}{t}x'(t) + x(t) = 0, t>0$. We know that $x(t) = \frac{\sin (t)}{t}$ is a solution of this ODE. Rewrite this second order ODE to a system of first order ODE's and find the fundamental matrix. Finally, determine the solution corresponding to the initial value's $x(\pi) = 1, x'(\pi) = 0$
My attempt
So first I rewrote the second order ODE to a system of first order ODE's: $$x''(t) + \frac{2}{t}x'(t) + x(t) = 0 \leftrightarrow x''(t) = -\frac{2}{t}x'(t) - x(t)$$ $$u(t) := \begin{pmatrix}x(t)\\x'(t)\end{pmatrix}, u'(t) := \begin{pmatrix}x'(t)\\x''(t)\end{pmatrix}$$ We get then the following system: $$u'(t) = \begin{pmatrix}0&1\\-1&-\frac{2}{t}\end{pmatrix}u(t)$$ I know have to find the fundamental matrix $U(t) = \begin{pmatrix}\frac{\sin (t)}{t}& y(t)\\\frac{\cos (t)}{t} - \frac{\sin (t)}{t^2}& y'(t)\end{pmatrix}$
Or in other words, I need to find $y$. Using the Wronskian of $U$ (which is $w(t) = \frac{w(t_0)t_0^2}{t}$ with $t_0$ being the initial value) I can determine that $$y'(t) = (\cot(t) - \frac{1}{t}) y(t) +\frac{w(t_0)t_0^2}{\sin (t)} $$ Since this is a linear ODE $y$ is explicitly calculated as $$y(t) = y(t_0)\exp\left(\int_{t_0}^t \cot (s) - \frac{1}{s} ds \right) + w(t_0)t_0^2\int_{t_0}^t \exp \left( \int_s^t \cot (r) - \frac{1}{r} dr\right)\frac{1}{\sin (s)} ds $$
But as far as I know, since $t_0 = \pi, \int_{\pi}^t \cot (s) ds$ is not defined.
The question is what is going wrong and/or what is supposed to be done to get $y(t)$?
I thank you in advance
Letting $x(t)=\frac{\sin t}{t}y(t)$ be a solution gives $$ \sin t y''+2\cos t y'=0. $$ This equation has solution $$ y=c_1+c_2\cot t. $$ From this, it is easy to get the fundamental matrix.