y^{''}+7y^'+6y=sin(2t),y(0)=1,y^'(0)=0

Lösung

y^{^{''}} + 7 y^{^{'}} + 6 y = sin (2 t), y (0) = 1, y^{^{'}} (0) = 0

y = (\frac{21}{100} + \frac{107}{100}) e^{- t} - \frac{21}{100} e^{- 6 t} + \frac{1}{100} sin (2 t) - \frac{7}{100} cos (2 t)

Schritte zur Lösung

y^{''} (t) + 7 y^{'} (t) + 6 y = sin (2 t)

Solve linear ODE:

y = (\frac{21}{100} + \frac{107}{100}) e^{- t} - \frac{21}{100} e^{- 6 t} + \frac{1}{100} sin (2 t) - \frac{7}{100} cos (2 t)

y = (\frac{21}{100} + \frac{107}{100}) e^{- t} - \frac{21}{100} e^{- 6 t} + \frac{1}{100} sin (2 t) - \frac{7}{100} cos (2 t)

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

y^{^{''}} + y^{^{'}} = e^{t} \cdot cos (t)

y^{^{''}} - (2 + 1)^{2} y = e^{28 t} + sin (2 + 1) t

y^{^{''}} - y^{^{'}} - 2 y = 2 x + 7

y^{^{''''}} + 12 y^{^{''}} + 36 = 0