y^{''}-10y^'+6y=0

Lösung

y^{^{''}} - 10 y^{^{'}} + 6 y = 0

y = c_{1} e^{(5 + 19) t} + c_{2} e^{(5 - 19) t}

Schritte zur Lösung

y^{''} (t) - 10 y^{'} (t) + 6 y = 0

Solve linear ODE:

y = c_{1} e^{(5 + 19) t} + c_{2} e^{(5 - 19) t}

y = c_{1} e^{(5 + 19) t} + c_{2} e^{(5 - 19) t}

y^{^{''}} + 8 y^{^{'}} + 41 y = 0, y (0) = 3, y^{^{'}} (0) = - 8

4 y^{^{''}} + 4 y^{^{'}} - 8 y = 0, y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} + 2 y^{^{'}} - 8 y = 0, y (0) = - 10, y^{^{'}} (0) = 10

l in e a r y^{^{''}} - y^{^{'}} - 2 y = 0

(y^{^{''}} + y)^{2} = 0