y^{''}+10y^'+288y=0

Lösung

y^{^{''}} + 10 y^{^{'}} + 288 y = 0

y = e^{- 5 t} (c_{1} cos (263 t) + c_{2} sin (263 t))

Schritte zur Lösung

y^{''} (t) + 10 y^{'} (t) + 288 y = 0

Solve linear ODE:

y = e^{- 5 t} (c_{1} cos (263 t) + c_{2} sin (263 t))

y = e^{- 5 t} (c_{1} cos (263 t) + c_{2} sin (263 t))

(d^{2} + 6) y = 0

y^{^{''}} + 9.82 \cdot y^{^{'}} = 0

y^{^{''}} - 2 y^{^{'}} - 3 y = 0, y (0) = 6, y^{^{'}} (0) = - 6

2 y^{^{''}} + 23 y^{^{'}} - 12 y = 0

a \frac{d ^{2} y}{d x ^{2}} - y b^{2} = 0