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

解答

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

y = c_{1} e^{4 x} + c_{2} e^{2 x} + \frac{1}{8} e^{- 2 x} + \frac{x}{4} + \frac{3}{16}

求解步骤

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

求解线性方程:

y = c_{1} e^{4 x} + c_{2} e^{2 x} + \frac{1}{8} e^{- 2 x} + \frac{x}{4} + \frac{3}{16}

y = c_{1} e^{4 x} + c_{2} e^{2 x} + \frac{1}{8} e^{- 2 x} + \frac{x}{4} + \frac{3}{16}

x^{^{''}} + 4 x^{^{'}} + 3 x = 1

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

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 t e^{2 t}, y (0) = 3, y^{^{'}} (0) = 0

y^{^{''}} + 36 y = 10 cos (6 x) + 15 sin (6 x)

25 y^{^{''}} + 20 y^{^{'}} + 4 y = 4 x^{2} - 3 x + 2