(D^5-2D^3-2D^2-3D-2)y=0

解答

(D^{5} - 2 D^{3} - 2 D^{2} - 3 D - 2) y = 0

y = c_{1} e^{2 x} + c_{2} e^{- x} + c_{3} x e^{- x} + c_{4} cos (x) + c_{5} sin (x)

求解步骤

(D^{5} - 2 D^{3} - 2 D^{2} - 3 D - 2) y = 0

求解线性方程:

y = c_{1} e^{2 x} + c_{2} e^{- x} + c_{3} x e^{- x} + c_{4} cos (x) + c_{5} sin (x)

y = c_{1} e^{2 x} + c_{2} e^{- x} + c_{3} x e^{- x} + c_{4} cos (x) + c_{5} sin (x)

y^{^{'''}} - 3 y^{^{''}} - y^{^{'}} + 3 y = 0

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = 0, y^{^{'}} (0) = 3

y^{^{''}} - 7 y = 0

(D^{5} - 15 D^{3} + 10 D^{2} + 60 D - 72) y = 0

y^{^{'''}} - 7 y^{^{''}} - 49 y^{^{'}} + 343 y = 0