Question: Use the method of reduction of order to find the general solution of the differential equation.
$(1+t^2)\frac{d^2y}{dt^2}-2t\frac{dy}{dt}+2y=0$ where $y_1(t)=t$
Here is my process:
Assume the 2nd solution will have the form:
$y_2(t)=v(t)y_1(t)$
We need:
$y_2(t)=vt$
$y'_2(t)=tv'+v$
$y''_2(t)=v''t+v'+v''$
Through substitution in the ode:
$(1+t^2)(v''t+v'+v'+v'')-2t(tv'+v)+2(vt)=0$
and by expanding:
$v''(t+1+t^2)+v'(1-t^2)=0$
So I change the variable:
$ w=v' \text{ and } w'=v'' $
Then:
$(t+1+t^2)w'+(1-t^2)w=0$
which is first order differential equation, and for some reason, I'm stuck at this part. I rewrote it as:
$(t+1+t^2)\frac{dw}{dt}+(1-t^2)w=0$
and I divided by $(t+1+t^2)$ to obtain:
$\frac{dw}{dt}+\frac{1-t^2}{t+1+t^2}w=0$
but I have no idea how to integrate: $\frac{1-t^2}{t+1+t^2}$, so I checked wolfram alpha and I'm pretty sure my professor would not give a complicated integral. So, I'm doing something wrong but I'm not sure.
