t^2*y'' - 2t*y' + y = 0
To get the solution do you need to do a reduction of order? If so how do you get one solution, trial and error?
Well, if you happen to stumble over a solution reduction of order will work well. Here's a general method.
Note that the order of the coefficient of $\displaystyle y^{(n)}$ is n. Thus this is an Euler differential equation.
So, make a change of variables:
$\displaystyle t = e^x \implies dt = e^x~dx \implies \frac{d}{dt} = e^{-x}\frac{d}{dx}$
And
$\displaystyle \frac{d^2}{dt^2} = \frac{d}{dt} \frac{d}{dt} = e^{-x}\frac{d}{dx} \left ( e^{-x} \frac{d}{dx} \right )$
$\displaystyle \frac{d^2}{dt^2} = e^{-x} \left ( -e^{-x} \frac{d}{dx} + e^{-x} \frac{d^2}{dx^2} \right )$
$\displaystyle \frac{d^2}{dt^2} = -e^{-2x} \frac{d}{dx} + e^{-2x} \frac{d^2}{dx^2}$
So the differential equation becomes:
$\displaystyle t^2*y^{\prime \prime} - 2t*y^{\prime} + y = 0$
$\displaystyle \left ( e^{x} \right )^2 \cdot \left (-e^{-2x} \frac{dy}{dx} + e^{-2x} \frac{d^2y}{dx^2} \right ) - 2e^x \cdot e^{-x}\frac{dy}{dx} + y = 0$
$\displaystyle \frac{d^2y}{dx^2} - 3 \frac{dy}{dx} + y = 0$
This is a linear homogeneous differential equation with constant coefficients and is easy to solve. So solve it and re-sub in $\displaystyle x = ln(t)$.
-Dan