y^{''}+4y^'=2y

Lösung

y^{^{''}} + 4 y^{^{'}} = 2 y

y = c_{1} e^{(- 2 + 6) t} + c_{2} e^{(- 2 - 6) t}

Schritte zur Lösung

y^{''} (t) + 4 y^{'} (t) = 2 y

Solve linear ODE:

y = c_{1} e^{(- 2 + 6) t} + c_{2} e^{(- 2 - 6) t}

y = c_{1} e^{(- 2 + 6) t} + c_{2} e^{(- 2 - 6) t}

(d^{4} - i^{2}) y = 0

(d^{3} - d^{2} + 2 d - d) y = 0

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

(d^{3} + d^{2} + d) \cdot y = 0

(d^{3} + d^{2} + d) y = 0