y^{'''}+2y^{''}-16y^'-32y=0

解答

y^{^{'''}} + 2 y^{^{''}} - 16 y^{^{'}} - 32 y = 0

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

求解步骤

y^{'''} (t) + 2 y^{''} (t) - 16 y^{'} (t) - 32 y = 0

求解线性方程:

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

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

x^{^{''}} - a^{2} x = 0

y^{^{''}} + 4 y^{^{'}} - 21 y = 0, y (0) = 1, y^{^{'}} (0) = 23

4 y^{^{''}} - 4 y^{^{'}} - 3 y = 0, y (0) = 1, y^{^{'}} (0) = 7

\frac{d ^{2} y}{d x ^{2}} + 81 y = 0

16 y^{^{''}} - 8 y^{^{'}} + 145 y = 0, y (0) = - 2, y^{^{'}} (0) = 1