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

Lösung

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

y = - \frac{1}{8} cos (t) + \frac{1}{8} sin (t) + \frac{1}{8} e^{- 2 t} sin (t) + \frac{1}{8} e^{- 2 t} cos (t)

Schritte zur Lösung

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

Solve linear ODE:

y = - \frac{1}{8} cos (t) + \frac{1}{8} sin (t) + \frac{1}{8} e^{- 2 t} sin (t) + \frac{1}{8} e^{- 2 t} cos (t)

y = - \frac{1}{8} cos (t) + \frac{1}{8} sin (t) + \frac{1}{8} e^{- 2 t} sin (t) + \frac{1}{8} e^{- 2 t} cos (t)

y^{^{''}} + 4 y^{^{'}} + 4 y = x^{3} e^{2 x}

y^{^{''}} - y^{^{'}} - 5 y = 5 t + 7

y^{^{''}} + 3 y = - 48 x e^{3 x}

y^{^{''}} + 8 y^{^{'}} + 8 y = cos (4 x)

y^{^{''}} + y^{^{'}} - 2 y = e^{x} sin (\frac{x}{2}) cos (\frac{x}{2})