y^{''}+2y^'+2y=\H(t-pi)

Lösung

y^{^{''}} + 2 y^{^{'}} + 2 y = H (t - π)

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) + e^{- t} cos (t) \cdot \int - e^{t} sin (t) H (t - π) d t + e^{- t} sin (t) \cdot \int e^{t} cos (t) H (t - π) d t

Schritte zur Lösung

y^{''} (t) + 2 y^{'} (t) + 2 y = H (t - π)

Solve linear ODE:

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) + e^{- t} cos (t) \cdot \int - e^{t} sin (t) H (t - π) d t + e^{- t} sin (t) \cdot \int e^{t} cos (t) H (t - π) d t

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) + e^{- t} cos (t) \cdot \int - e^{t} sin (t) H (t - π) d t + e^{- t} sin (t) \cdot \int e^{t} cos (t) H (t - π) d t

d (x) = \frac{- 6}{- 3 x + 4}

y^{^{'''}} - 3 y^{^{''}} + 2 y^{^{'}} = x

y^{^{''}} - 6 y^{^{'}} + 8 y^{^{'}} = x^{2} - 1

y^{^{''}} + 2 y^{^{'}} + y = 4 e^{- 2 t} + t + t^{2}

D^{3} y = e^{3 x}