y^{''}+40y^'+4000y=0

Lösung

y^{^{''}} + 40 y^{^{'}} + 4000 y = 0

y = e^{- 20 t} (c_{1} cos (60 t) + c_{2} sin (60 t))

Schritte zur Lösung

y^{''} (t) + 40 y^{'} (t) + 4000 y = 0

Solve linear ODE:

y = e^{- 20 t} (c_{1} cos (60 t) + c_{2} sin (60 t))

y = e^{- 20 t} (c_{1} cos (60 t) + c_{2} sin (60 t))

- 5 y^{^{'}} + 2 y^{^{''}} - y^{^{'}} - 6 y = 0

l in e a r y^{^{''}} + 4 y = 0, y (0) = 2, y^{^{'}} (0) = 2

49 y^{^{''}} - 126 y^{^{'}} + 117 y = 0, y (4) = 2, y^{^{'}} (4) = 2

y^{^{'''}} - 3 y^{^{''}} - 49 y^{^{'}} - 45 y = 0

y^{^{''}} + 5 y^{^{'}} + 6 y = 0, y (0) = 6, y^{^{'}} (0) = - 8