y^{'''}+y^{''}-22y^'-40y=0

解答

y^{^{'''}} + y^{^{''}} - 22 y^{^{'}} - 40 y = 0

y = c_{1} e^{- 2 t} + c_{2} e^{- 4 t} + c_{3} e^{5 t}

求解步骤

y^{'''} (t) + y^{''} (t) - 22 y^{'} (t) - 40 y = 0

求解线性方程:

y = c_{1} e^{- 2 t} + c_{2} e^{- 4 t} + c_{3} e^{5 t}

y = c_{1} e^{- 2 t} + c_{2} e^{- 4 t} + c_{3} e^{5 t}

y^{^{''}} + 11 y^{^{'}} + 10 y = 0

81 y^{^{''}} - 144 y^{^{'}} + 73 y = 0, y (1) = 6, y^{^{'}} (1) = 5

y^{^{''}} + 9 y^{^{'}} + 8 y = 0, y (0) = 6, y^{^{'}} (0) = - 1

y^{^{''''}} + 4 y^{^{'''}} - y^{^{''}} - 16 y^{^{'}} - 12 y = 0

x^{^{'''}} + 3 x^{^{''}} + 3 x^{^{'}} + x = 0