x^{''}+4x^'+8x=0

Lösung

x^{^{''}} + 4 x^{^{'}} + 8 x = 0

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

Schritte zur Lösung

x^{''} (t) + 4 x^{'} (t) + 8 x = 0

Solve linear ODE:

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

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

9 y^{^{''}} + 25 y = 0

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

49 y^{^{''}} - 168 y^{^{'}} + 144 y = 0

(D^{3} - 8) y = 0

4 \frac{d ^{2} y}{d x ^{2}} + 36 y = 0