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

解

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

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) + e^{x} - \frac{6}{5} sin (x) + \frac{3}{5} cos (x)

解答ステップ

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

線形 ODE を解く:

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) + e^{x} - \frac{6}{5} sin (x) + \frac{3}{5} cos (x)

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) + e^{x} - \frac{6}{5} sin (x) + \frac{3}{5} cos (x)