(0.1)x^{''}+(0.4)x^'+2x=0

解答

(0.1) x^{^{''}} + (0.4) x^{^{'}} + 2 x = 0

x = e^{- 2 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

求解步骤

(0.1) x^{''} (t) + (0.4) x^{'} (t) + 2 x = 0

求解线性方程:

x = e^{- 2 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

x = e^{- 2 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

10 x^{^{''}} + 40 x^{^{'}} + \frac{1}{50} x = 0

y^{^{''}} + 2 y^{^{'}} + 26 y = 0, y (0) = 3, y^{^{'}} (0) = - 1

y^{^{''}} - 12 y^{^{'}} + 52 y = 0, y (\frac{π}{8}) = 1, y (0) = 8

y^{^{''}} + y = 0, y^{^{'}} (π) = 1, y (0) = 1

y^{^{'''}} - 2 y^{^{''}} - 29 y^{^{'}} - 42 y = 0