y^{''}-4y^'+8y=x^3

解答

y^{^{''}} - 4 y^{^{'}} + 8 y = x^{3}

y = e^{2 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{x ^{3}}{8} + \frac{3 x ^{2}}{16} + \frac{3 x}{32}

求解步骤

y^{''} (x) - 4 y^{'} (x) + 8 y = x^{3}

求解线性方程:

y = e^{2 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{x ^{3}}{8} + \frac{3 x ^{2}}{16} + \frac{3 x}{32}

y = e^{2 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{x ^{3}}{8} + \frac{3 x ^{2}}{16} + \frac{3 x}{32}

y^{^{''}} + 4 y = x^{2} + e^{- x}

2 y^{^{''}} - 7 y^{^{'}} - 4 y = 7 - 2 x

y^{^{''}} + 3 y^{^{'}} = sin (3 x)

y^{^{''}} + 16 y = 5 sin (x)

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