3y^{''}+4y^'+5y=0

Lösung

3 y^{^{''}} + 4 y^{^{'}} + 5 y = 0

y = e^{- \frac{2 t}{3}} (c_{1} cos (\frac{11 t}{3}) + c_{2} sin (\frac{11 t}{3}))

Schritte zur Lösung

3 y^{''} (t) + 4 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = e^{- \frac{2 t}{3}} (c_{1} cos (\frac{11 t}{3}) + c_{2} sin (\frac{11 t}{3}))

y = e^{- \frac{2 t}{3}} (c_{1} cos (\frac{11 t}{3}) + c_{2} sin (\frac{11 t}{3}))

y^{^{''}} - 5 y^{^{''}} + 24 y^{^{'}} - 20 y = 0

y^{^{''''}} + 8 y^{^{'''}} + 21 y^{^{''}} + 22 y^{^{'}} + 8 y = 0

y^{^{''}} + 9 y^{^{'}} + α y = 0, y (0) = 1, y^{^{'}} (0) = - 4

y^{^{''}} + 0.05 y^{^{'}} - 0.0006 y = 0

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = - 1