x^{''}+2x+3x^'=5cos(t)+sin(2t)

解答

x^{^{''}} + 2 x + 3 x^{^{'}} = 5 cos (t) + sin (2 t)

x = c_{1} e^{- t} + c_{2} e^{- 2 t} + \frac{3}{2} sin (t) + \frac{1}{2} cos (t) - \frac{1}{20} sin (2 t) - \frac{3}{20} cos (2 t)

求解步骤

x^{''} (t) + 2 x + 3 x^{'} (t) = 5 cos (t) + sin (2 t)

求解线性方程:

x = c_{1} e^{- t} + c_{2} e^{- 2 t} + \frac{3}{2} sin (t) + \frac{1}{2} cos (t) - \frac{1}{20} sin (2 t) - \frac{3}{20} cos (2 t)

x = c_{1} e^{- t} + c_{2} e^{- 2 t} + \frac{3}{2} sin (t) + \frac{1}{2} cos (t) - \frac{1}{20} sin (2 t) - \frac{3}{20} cos (2 t)

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

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

y^{^{''}} + 10 y^{^{'}} + 61 y = 5 cos (7 t)

(D^{4} - 1) y = 7 x^{2}

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e^{5 x}