y^{''}-2y^'+5y=e^x(0=(2x))

解答

y^{^{''}} - 2 y^{^{'}} + 5 y = e^{x} (0 = (2 x))

y = e^{x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{2 x}{5} + \frac{4}{25}

求解步骤

y^{''} (x) - 2 y^{'} (x) + 5 y = e^{x} (0 = (2 x))

求解线性方程:

y = e^{x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{2 x}{5} + \frac{4}{25}

y = e^{x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{2 x}{5} + \frac{4}{25}

y^{^{''}} - 6 y^{^{'}} + 9 y = 18

y^{^{''}} - y = 8 t e^{t} + 2 e^{t}

y^{^{''}} + 18 y = 3 e^{8 x}

y^{^{''}} - 4 y^{^{'}} + 4 y = \frac{e ^{2 x}}{1 + x}

q^{^{''}} + 3 q^{^{'}} + 2 q = e^{- t} + e^{- 2 t}