5y^{''}+25y=0

Lösung

5 y^{^{''}} + 25 y = 0

y = c_{1} cos (5 t) + c_{2} sin (5 t)

Schritte zur Lösung

5 y^{''} (t) + 25 y = 0

Solve linear ODE:

y = c_{1} cos (5 t) + c_{2} sin (5 t)

y = c_{1} cos (5 t) + c_{2} sin (5 t)

- x^{^{''}} + 2 x^{^{'}} + x = 0

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

(D^{2} - 1) y = 0

9 y^{^{''}} - 5 y^{^{'}} - 4 y = 0

y^{^{''}} + 16 y = 0, y (0) = 2, y (0) = - 2