If the differential equation $t^2 y'' - 2y' + (3 + t)y = 0$ has $y_1$ and $y_2$ as a fundamental set of solutions and if $W(y_1, y_2)(2) = 3$, find $W(y_1, y_2)(4)$.
Is it possible for me to solve this problem as such:
If $W(y_1, y_2)(2) = 3$,
then $W(y_1, y_2)(4) = W(y_1, y_2)(2^2) = 3^2$.
Therefore, $W(y_1, y_2)(4) = 9$
I'm not sure if this is an acceptable way to solve this question or not, and if it's not, could someone please explain why it would be wrong, and how I could go about solving it correctly?
You know that for $W=\det\pmatrix{y_1&y_2\\y_1'&y_2'}$ you get $$ W'=\det\pmatrix{y_1&y_2\\y_1''&y_2''}=\frac2{t^2}W $$ so that $W(t)=Ce^{-2/t}=3e^{1-2/t}$. So no, your solution for $W(4)$ is wrong.