y^{''}+32y^'+160y=0

Lösung

y^{^{''}} + 32 y^{^{'}} + 160 y = 0

y = c_{1} e^{4 (6 - 4) t} + c_{2} e^{- 4 (4 + 6) t}

Schritte zur Lösung

y^{''} (t) + 32 y^{'} (t) + 160 y = 0

Solve linear ODE:

y = c_{1} e^{4 (6 - 4) t} + c_{2} e^{- 4 (4 + 6) t}

y = c_{1} e^{4 (6 - 4) t} + c_{2} e^{- 4 (4 + 6) t}

y^{^{''}} - 256 y = 0

(D^{3} - 6 D^{2} + 5 D) y = 0

y^{^{''''}} - 6 y^{^{''}} + 9 y = 0

y^{^{'''}} + y^{^{''}} - 72 y^{^{'}} = 0

y^{^{'''}} + 4 y^{^{''}} + 13 y^{^{'}} = 0