# Use the method of reduction of order to find the general solution of the differential equation.

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.