y^{''}-4y^'+8y=e^{2x}+sin(2x)

解

y^{^{''}} - 4 y^{^{'}} + 8 y = e^{2 x} + sin (2 x)

y = e^{2 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{1}{4} e^{2 x} + \frac{1}{20} sin (2 x) + \frac{1}{10} cos (2 x)

解答ステップ

y^{''} (x) - 4 y^{'} (x) + 8 y = e^{2 x} + sin (2 x)

線形 ODE を解く:

y = e^{2 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{1}{4} e^{2 x} + \frac{1}{20} sin (2 x) + \frac{1}{10} cos (2 x)

y = e^{2 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{1}{4} e^{2 x} + \frac{1}{20} sin (2 x) + \frac{1}{10} cos (2 x)