y^{''}-2y+5y=0

Lösung

y^{^{''}} - 2 y + 5 y = 0

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

Schritte zur Lösung

y^{''} (t) - 2 y + 5 y = 0

Solve linear ODE:

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

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

y^{^{''}} - 12 y^{^{'}} + 27 y = 0

y^{^{'''}} - 4 y^{^{''}} + 8 y^{^{'}} = 0

6 y^{^{''}} + 11 y^{^{'}} - 6 y = 0

(D^{2} - D - 6) y = 0

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