Question

\[t^2 y''-4 t y'+6y==0\]

Answer 1

\[y_1=t^2\]

Answer 2

\[y_2= v(t) y_1\]

\[y_2= v(t) t^2\]

\[y'=v' t^2+v 2t\]

\[y''=4v' t+v'' t^2+ 2v\]

plugging into differential equation, we get

\[v''(t) t^4=0\]

\[v'(t) =D\]

\[v(t) =Dt+C\]


\[y_2= v t^2\]

\[y_2 =t^3\]