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

Lösung

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)

Schritte zur Lösung

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

Solve linear ODE:

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)

y^{^{''}} + y^{^{'}} - 2 y = - 260 sin (3 t)

y^{^{''}} - 9 y = 81 x^{3}

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

y^{^{''}} + 2 y^{^{'}} + y = 2 x + 5 - e^{- 2 x}

y^{^{''}} - y^{^{'}} - 2 y = 4 t^{2}, y (0) = 1, y^{^{'}} (0) = 4