y^{''}-6y^'+8y=-3e^{2x}-5xe^{-4x}+3x

解答

y^{^{''}} - 6 y^{^{'}} + 8 y = - 3 e^{2 x} - 5 x e^{- 4 x} + 3 x

y = c_{1} e^{4 x} + c_{2} e^{2 x} + \frac{3 e ^{2 x} x}{2} - \frac{5 e ^{- 4 x} x}{48} - \frac{35}{1152} e^{- 4 x} + \frac{3 x}{8} + \frac{9}{32}

求解步骤

y^{''} (x) - 6 y^{'} (x) + 8 y = - 3 e^{2 x} - 5 x e^{- 4 x} + 3 x

求解线性方程:

y = c_{1} e^{4 x} + c_{2} e^{2 x} + \frac{3 e ^{2 x} x}{2} - \frac{5 e ^{- 4 x} x}{48} - \frac{35}{1152} e^{- 4 x} + \frac{3 x}{8} + \frac{9}{32}

y = c_{1} e^{4 x} + c_{2} e^{2 x} + \frac{3 e ^{2 x} x}{2} - \frac{5 e ^{- 4 x} x}{48} - \frac{35}{1152} e^{- 4 x} + \frac{3 x}{8} + \frac{9}{32}

y^{^{''''}} - y^{^{''}} = 3 x^{2} - 2 x

y^{^{'''}} - 2 y^{^{''}} + y^{^{'}} = 1

\frac{1}{6} y^{^{''}} + 6 y = tan (6 x) - \frac{1}{3} e^{3 x}

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

y^{^{'''}} - y^{^{'}} = 10 cos (2 x)