y^{''}+20y=0

Lösung

y^{^{''}} + 20 y = 0

y = c_{1} cos (25 t) + c_{2} sin (25 t)

Schritte zur Lösung

y^{''} (t) + 20 y = 0

Solve linear ODE:

y = c_{1} cos (25 t) + c_{2} sin (25 t)

y = c_{1} cos (25 t) + c_{2} sin (25 t)

(D^{2} + 4 D + 5) y = 0

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

y^{^{''''}} + 4 y^{^{'''}} + 6 y^{^{''}} + 4 y^{^{'}} + y = 0

11 x^{^{''}} + 112.5 x = 0

x^{^{''}} + 3 x^{^{'}} = 0