Question

Diff EQ variation of parameters problem, i think my V1 is wrong, but idk. see attached picture

Answer 1

That first bit doesn't work because the coefficients aren't constant. The solutions to
\[t^2 y'' - ty' =0\]
are polynomials. Note: if
\[y = t^n\]
then
\[y' = nt^{n-1}, y'' = n(n-1)t^{n-2}\]
so that equation gives
\[(n(n-1)-n)t=0\]
so
\[n^2-2n = 0 \rightarrow n = 0, 2\]
and the homogeneous solution is
\[y_h = c_1t^2 + c_2\]

Answer 2

ohh that makes sense. Thank you