x^{''}+6x^'+9x=10cos(2t)

解答

x^{^{''}} + 6 x^{^{'}} + 9 x = 10 cos (2 t)

x = c_{1} e^{- 3 t} + c_{2} t e^{- 3 t} + \frac{120}{169} sin (2 t) + \frac{50}{169} cos (2 t)

求解步骤

x^{''} (t) + 6 x^{'} (t) + 9 x = 10 cos (2 t)

求解线性方程:

x = c_{1} e^{- 3 t} + c_{2} t e^{- 3 t} + \frac{120}{169} sin (2 t) + \frac{50}{169} cos (2 t)

x = c_{1} e^{- 3 t} + c_{2} t e^{- 3 t} + \frac{120}{169} sin (2 t) + \frac{50}{169} cos (2 t)

y^{^{'''}} + 2 y^{^{'}} = 3 (y^{^{''}} - x^{2} + 2)

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

y^{^{'''}} - 3 y^{^{'}} - 2 y = e^{x} (x e^{x} + 1)

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 y = e^{x}

y^{^{''}} + 2 y^{^{'}} + 2 y = x e^{- x}