Question

Solve \(y^{(4)} - 8y''' + 26y'' - 40y' + 25y =0 \)

Answer 1

\[y^{(4)} - 8y''' + 26y'' - 40y' + 25y =0 \]
\[λ^4 - 8λ^3 + 26λ^2 -40λ + 25 =0 \]
\[(λ^2 -4λ + 5)^2 = 0\]
λ = (2+i) (repeated) or λ = 2-i (repeated)

\[y = c_{1} e^{(2+i)x} + c_{2} x e^{(2+i)x} + c_{3} e^{(2-i)x} + c_{4} xe^{(2-i)x} \]\[= e^{2x} (c_{1} e^{ix} + c_{2} x e^{ix} + c_{3} e^{-ix} + c_{4} xe^{-ix})\]\[= e^{2x} [c_{1} e^{ix} + c_{3} e^{-ix} + x(c_{2} e^{ix} + c_{4} e^{-ix})]\]\[= e^{2x} [c_{1} (cosx + isinx) + c_{3} (cos(-x)+i(sin(-x)) \]\[\ \ \ \ \ \ + x(c_{2} (cosx + isinx) + c_{4} (cos(-x)+i(sin(-x)))]\]\[= e^{2x} [(Acosx + Bsinx) + x(Ccosx + Dsinx) \]

Answer 2

The last line should be
\[= e^{2x} [(Acosx + Bsinx) + x(Ccosx + Dsinx)]\]

Answer 3

so far ... seems correct!!

Answer 4

:)