y^{''}-6y^'+32y=0

Lösung

y^{^{''}} - 6 y^{^{'}} + 32 y = 0

y = e^{3 t} (c_{1} cos (23 t) + c_{2} sin (23 t))

Schritte zur Lösung

y^{''} (t) - 6 y^{'} (t) + 32 y = 0

Solve linear ODE:

y = e^{3 t} (c_{1} cos (23 t) + c_{2} sin (23 t))

y = e^{3 t} (c_{1} cos (23 t) + c_{2} sin (23 t))

y^{^{''}} + 2 y^{^{'}} - 15 y = 0, y (0) = 6, y^{^{'}} (0) = - 6

y^{^{''}} - 12 y^{^{'}} + 45 y = 0, y (0) = 5, y^{^{'}} (0) = - 3

10 y^{^{''}} + 45 y^{^{'}} + 2500 y = 0

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (- 1) = 3, y^{^{'}} (- 1) = 6

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (- 1) = 3, y^{^{'}} (- 1) = 5