y^{''}+2y^'+y=sin(x)+2cos(2x)

解答

y^{^{''}} + 2 y^{^{'}} + y = sin (x) + 2 cos (2 x)

y = c_{1} e^{- x} + c_{2} x e^{- x} - \frac{1}{2} cos (x) + \frac{8}{25} sin (2 x) - \frac{6}{25} cos (2 x)

求解步骤

y^{''} (x) + 2 y^{'} (x) + y = sin (x) + 2 cos (2 x)

求解线性方程:

y = c_{1} e^{- x} + c_{2} x e^{- x} - \frac{1}{2} cos (x) + \frac{8}{25} sin (2 x) - \frac{6}{25} cos (2 x)

y = c_{1} e^{- x} + c_{2} x e^{- x} - \frac{1}{2} cos (x) + \frac{8}{25} sin (2 x) - \frac{6}{25} cos (2 x)

(d^{2} - 1) y = x^{2}

y^{^{'''}} - 9 y^{^{''}} + 27 y^{^{'}} - 27 y = 4 e^{3 x}

(D^{3} - 7 D - 6) y = e^{2 x} (1 + x)

y^{^{''}} - 2 y^{^{'}} - 3 y = 8 e^{5 x}

y^{^{''}} + 8 y = 5 x + 2 e^{- x}