y^{''}+2y^'+4y=0,y(0)=1

Lösung

y^{^{''}} + 2 y^{^{'}} + 4 y = 0, y (0) = 1

y = e^{- t} (cos (3 t) + c_{2} sin (3 t))

Schritte zur Lösung

y^{''} (t) + 2 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = e^{- t} (cos (3 t) + c_{2} sin (3 t))

y = e^{- t} (cos (3 t) + c_{2} sin (3 t))

y^{^{''}} + 2 y^{^{'}} + 1 y = 0, y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 3, y^{^{'}} (0) = - 7

y^{^{''''}} - 4 y^{^{'''}} + 8 y^{^{''}} - 16 y^{^{'}} + 16 y = 0

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 3, y^{^{'}} (0) = - 2

2 y^{^{''}} - 2 y^{^{'}} + y = 0, y (0) = - 1, y^{^{'}} (0) = 2