y^{'''}-5y^{''}-25y^'+125y=0

解答

y^{^{'''}} - 5 y^{^{''}} - 25 y^{^{'}} + 125 y = 0

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

求解步骤

y^{'''} (t) - 5 y^{''} (t) - 25 y^{'} (t) + 125 y = 0

求解线性方程:

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

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

y^{^{''}} - y^{^{'}} - 12 y = 0, y (1) = 0, y^{^{'}} (1) = 1

y^{^{''}} - 2 y^{^{'}} - 3 y = 0, y (0) = 3, y^{^{'}} (0) = 2

\frac{d ^{2} y}{d x ^{2}} - 8 \frac{d y}{d x} + 15 y = 0

y^{^{''}} - 11 y^{^{'}} + 24 y = 0

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