y^{''}+4y^1-y=0

Lösung

y^{^{''}} + 4 y^{1} - y = 0

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

Schritte zur Lösung

y^{''} (t) + 4 y^{1} - 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^{^{''}} + π^{2} \cdot y = 0

y^{^{''}} + (\frac{25}{6}) y^{^{'}} + 25 y = 0

(D^{3} - 1) y = 0

28 y^{^{''}} + 17 y^{^{'}} - 3 y = 0

x^{^{''}} + 6 x^{^{'}} + 25 x = 0, x (0) = \frac{1}{2}, x^{^{'}} (0) = 0