y^{''}-2y^'+2y=e^2x(cos(x)-8sin(x))

解答

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{2} x (cos (x) - 8 sin (x))

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) - 2 e^{2} x sin (x) - 3 e^{2} x cos (x) + (\frac{44 e ^{2}}{5} - 10 e^{2}) sin (x) - \frac{22 e ^{2}}{5} cos (x)

求解步骤

y^{''} (x) - 2 y^{'} (x) + 2 y = e^{2} x (cos (x) - 8 sin (x))

求解线性方程:

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) - 2 e^{2} x sin (x) - 3 e^{2} x cos (x) + (\frac{44 e ^{2}}{5} - 10 e^{2}) sin (x) - \frac{22 e ^{2}}{5} cos (x)

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) - 2 e^{2} x sin (x) - 3 e^{2} x cos (x) + (\frac{44 e ^{2}}{5} - 10 e^{2}) sin (x) - \frac{22 e ^{2}}{5} cos (x)

y^{^{''}} - 4 y = 6 cos (3 x + 3 0^{\circ})

y^{^{'''}} + y^{^{''}} + y^{^{'}} = e^{x}

y^{^{''}} - 8 y^{^{'}} + 16 y = 2 e^{4 x} + 3 cos (4 x) - 2 x + 1

x^{^{''}} - 6 x^{^{'}} + 9 x = 5 t e^{3 t}

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