y^{''''}+16y=0

Lösung

y^{^{''''}} + 16 y = 0

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

Schritte zur Lösung

y^{(4)} (t) + 16 y = 0

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 3, y^{^{'}} (0) = 5

y^{^{''}} + n^{2} y = 0

l in e a r \frac{d ^{2} y}{d x ^{2}} + 5 \frac{d y}{d x} + 10 y = 0

θ^{^{''}} (t) + \frac{4 k}{3} θ (t) = 0

y^{^{''}} + 144 y = 0, y (0) = 3, y^{^{'}} (0) = - 48