y^{'''}-2y^'+4y=0

解答

y^{^{'''}} - 2 y^{^{'}} + 4 y = 0

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

求解步骤

y^{'''} (t) - 2 y^{'} (t) + 4 y = 0

求解线性方程:

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

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

\frac{1}{4} y^{^{''}} + 2 y^{^{'}} + 2 y = 0

4 y^{^{''}} - 16 y^{^{'}} + 17 y = 0

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = 0, y (0) = 3

y^{^{''}} + (a - 1) y^{^{'}} + b y = 0

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