though theres some fiddling since the final two terms arise from complex roots. In any case, the first two roots are 2 and 1/4. The second two arise from -1 + i and -1 - i. The characteristic equation is then(m - 2)(m - 1/4)(m - (-1 + i))(m - (-1 - i))= [4m^4 - m^3 - 8m^2 - 14m + 4]/4we can multiply by 4 and the underlying DE doesnt change, giving4y - y - 8y - 14y + 4 = 0Showing linear independence can in fact be done with the Wronskian.......
2, 1/4, -1 ± i.
Can you see where I got these?
Anyway, as for the follow up question: These could NOT be a fundamental solution set to a fifth order equation. That would require 5 linearly independent functions.
Suppose they are linearly independent (they are). The solutions you have essentially give you the roots of the characteristic equation, though there's some fiddling since the final two terms arise from complex roots. In any case, the first two roots are 2 and 1/4. The second two arise from -1 + i and -1 - i. The characteristic equation is then
(m - 2)(m - 1/4)(m - (-1 + i))(m - (-1 - i))
= [4m^4 - m^3 - 8m^2 - 14m + 4]/4
we can multiply by 4 and the underlying DE doesn't change, giving
4y'''' - y''' - 8y'' - 14y' + 4 = 0
Showing linear independence can in fact be done with the Wronskian. You can skip computing the determinant explicitly by using the Vandermonde determinant identity [...on the transpose...]. This method generalizes, actually: as long as the roots are distinct, the induced fundamental set is linearly independent.
