y^{''}+1/8 y^'+4y=3cos(6t)

解

y^{^{''}} + \frac{1}{8} y^{^{'}} + 4 y = 3 cos (6 t)

y = e^{- \frac{t}{16}} (c_{1} cos (\frac{1023 t}{16}) + c_{2} sin (\frac{1023 t}{16})) + \frac{36}{16393} sin (6 t) - \frac{1536}{16393} cos (6 t)

解答ステップ

y^{''} (t) + \frac{1}{8} y^{'} (t) + 4 y = 3 cos (6 t)

線形 ODE を解く:

y = e^{- \frac{t}{16}} (c_{1} cos (\frac{1023 t}{16}) + c_{2} sin (\frac{1023 t}{16})) + \frac{36}{16393} sin (6 t) - \frac{1536}{16393} cos (6 t)

y = e^{- \frac{t}{16}} (c_{1} cos (\frac{1023 t}{16}) + c_{2} sin (\frac{1023 t}{16})) + \frac{36}{16393} sin (6 t) - \frac{1536}{16393} cos (6 t)