2x^{''}+8x^'+52x=0

Lösung

2 x^{^{''}} + 8 x^{^{'}} + 52 x = 0

x = e^{- 2 t} (c_{1} cos (22 t) + c_{2} sin (22 t))

Schritte zur Lösung

2 x^{''} (t) + 8 x^{'} (t) + 52 x = 0

Solve linear ODE:

x = e^{- 2 t} (c_{1} cos (22 t) + c_{2} sin (22 t))

x = e^{- 2 t} (c_{1} cos (22 t) + c_{2} sin (22 t))

y^{^{'''}} + 4 y^{^{''}} - 7 y^{^{'}} - 10 y = 0

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (0) = 5, y^{^{'}} (0) = 0

(y^{^{''''}} + 8 y^{^{''}} + 16 y)^{2} = 0

y^{^{''}} + 5 y^{^{'}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = 1