y^{''}+14y^'+51y=0

Lösung

y^{^{''}} + 14 y^{^{'}} + 51 y = 0

y = e^{- 7 t} (c_{1} cos (2 t) + c_{2} sin (2 t))

Schritte zur Lösung

y^{''} (t) + 14 y^{'} (t) + 51 y = 0

Solve linear ODE:

y = e^{- 7 t} (c_{1} cos (2 t) + c_{2} sin (2 t))

y = e^{- 7 t} (c_{1} cos (2 t) + c_{2} sin (2 t))

y^{^{'''}} - 4 y^{^{''}} - 5 y^{^{'}} = 0, y (1) = 0, y^{^{'}} (1) = 8

y^{^{''}} - 1 y^{^{'}} + 0.1 y = 0

(d^{4} - 1) y = 0

4 y^{^{''}} - 2 y^{^{'}} - 2 y = 0, y (0) = 2, y^{^{'}} (0) = 1

u^{^{''}} + 49 H (t) = 0, H (0) = 3, u^{^{'}} (0) = - 7