y^{''}+4y=-4y^'

Lösung

y^{^{''}} + 4 y = - 4 y^{^{'}}

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t}

Schritte zur Lösung

y^{''} (t) + 4 y = - 4 y^{'} (t)

Solve linear ODE:

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t}

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t}

so l v e f or u, \frac{d ^{2} u}{d x ^{2}} = u

16 y^{^{''}} + 8 y^{^{'}} + 5 y = 0

y^{^{''}} + 20 y = 0

(D^{2} + 4 D + 5) y = 0

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