y^{''}+10y^'+2y=0

Lösung

y^{^{''}} + 10 y^{^{'}} + 2 y = 0

y = c_{1} e^{(- 5 + 23) t} + c_{2} e^{(- 5 - 23) t}

Schritte zur Lösung

y^{''} (t) + 10 y^{'} (t) + 2 y = 0

Solve linear ODE:

y = c_{1} e^{(- 5 + 23) t} + c_{2} e^{(- 5 - 23) t}

y = c_{1} e^{(- 5 + 23) t} + c_{2} e^{(- 5 - 23) t}

x^{^{''}} + 5 x^{^{'}} + 3 x = 0, x (0) = 5, x^{^{'}} (0) = 7

y^{^{''}} - 22 y^{^{'}} + 121 y = 0, y (0) = 0, y^{^{'}} (0) = 5

y^{^{''}} + 9 y^{^{'}} + 14 y = 0, y (0) = 0, y^{^{'}} (0) = - 1

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = - 9

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