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

解答

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

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{3}{5} e^{2 x} sin (x) - \frac{1}{5} e^{2 x} cos (x)

求解步骤

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

求解线性方程:

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{3}{5} e^{2 x} sin (x) - \frac{1}{5} e^{2 x} cos (x)

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{3}{5} e^{2 x} sin (x) - \frac{1}{5} e^{2 x} cos (x)

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

y^{^{''}} + 100 y^{^{'}} + \frac{1}{0.0004} y = 30, y (0) = 0

y^{^{''''}} - 5 y^{^{'''}} + 4 y^{^{''}} = x

2 y^{^{''}} + y = 5 + cos (4 t)

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