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

Lösung

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

y = c_{1} + c_{2} e^{- t} + e^{\frac{t}{2}} (c_{3} cos (\frac{3 t}{2}) + c_{4} sin (\frac{3 t}{2}))

Schritte zur Lösung

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- t} + e^{\frac{t}{2}} (c_{3} cos (\frac{3 t}{2}) + c_{4} sin (\frac{3 t}{2}))

y = c_{1} + c_{2} e^{- t} + e^{\frac{t}{2}} (c_{3} cos (\frac{3 t}{2}) + c_{4} sin (\frac{3 t}{2}))

y^{^{''}} + 4 y^{^{'}} + 20 y = 0, y (0) = 2, y^{^{'}} (0) = - 8

(y^{^{''}} - 6 y^{^{'}} + 9 y)^{3} = 0

y^{^{''''}} + 4 y^{^{'''}} - 3 y^{^{''}} - 3 y^{^{'}} + \frac{1}{2} y = 0

(d^{3} - 6 d^{2} + 11 d - 6) y = 0

8 y^{^{''''}} + 6 y^{^{'''}} + 33 y^{^{''}} + 24 y^{^{'}} + 4 y = 0