y^{''''}-3y^{'''}-6y^{''}+28y^'-24y=0

解答

y^{^{''''}} - 3 y^{^{'''}} - 6 y^{^{''}} + 28 y^{^{'}} - 24 y = 0

y = c_{1} e^{- 3 t} + c_{2} e^{2 t} + c_{3} t e^{2 t} + c_{4} t^{2} e^{2 t}

求解步骤

y^{(4)} (t) - 3 y^{'''} (t) - 6 y^{''} (t) + 28 y^{'} (t) - 24 y = 0

求解线性方程:

y = c_{1} e^{- 3 t} + c_{2} e^{2 t} + c_{3} t e^{2 t} + c_{4} t^{2} e^{2 t}

y = c_{1} e^{- 3 t} + c_{2} e^{2 t} + c_{3} t e^{2 t} + c_{4} t^{2} e^{2 t}

5 y^{^{''}} + 20 y^{^{'}} + 10 y = 0

l in e a r y^{^{''}} + (\frac{216}{7}) y = 0

(D^{5} - 5 D^{4} + 7 D^{3} + D^{2} - 8 D + 4) y = 0

y^{^{''}} - 7 y + 12 y = 0

(y^{^{''}} + 2 y^{^{'}} + 5 y) = 0