y^{''}+2y^'+2y=sin(t),y(0)=0,y^'(0)=0

解

y^{^{''}} + 2 y^{^{'}} + 2 y = sin (t), y (0) = 0, y^{^{'}} (0) = 0

y = e^{- t} (\frac{2}{5} cos (t) + \frac{1}{5} sin (t)) + \frac{1}{5} sin (t) - \frac{2}{5} cos (t)

解答ステップ

y^{''} (t) + 2 y^{'} (t) + 2 y = sin (t)

線形 ODE を解く:

y = e^{- t} (\frac{2}{5} cos (t) + \frac{1}{5} sin (t)) + \frac{1}{5} sin (t) - \frac{2}{5} cos (t)

y = e^{- t} (\frac{2}{5} cos (t) + \frac{1}{5} sin (t)) + \frac{1}{5} sin (t) - \frac{2}{5} cos (t)