2y^{''}+6y^'+10y=4cos(t)

解

2 y^{^{''}} + 6 y^{^{'}} + 10 y = 4 cos (t)

y = e^{- \frac{3 t}{2}} (c_{1} cos (\frac{11 t}{2}) + c_{2} sin (\frac{11 t}{2})) + \frac{6}{25} sin (t) + \frac{8}{25} cos (t)

解答ステップ

2 y^{''} (t) + 6 y^{'} (t) + 10 y = 4 cos (t)

線形 ODE を解く:

y = e^{- \frac{3 t}{2}} (c_{1} cos (\frac{11 t}{2}) + c_{2} sin (\frac{11 t}{2})) + \frac{6}{25} sin (t) + \frac{8}{25} cos (t)

y = e^{- \frac{3 t}{2}} (c_{1} cos (\frac{11 t}{2}) + c_{2} sin (\frac{11 t}{2})) + \frac{6}{25} sin (t) + \frac{8}{25} cos (t)