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

Lösung

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

y = c_{1} + c_{2} e^{- 4 t} + \frac{t}{2}

Schritte zur Lösung

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 t} + \frac{t}{2}

y = c_{1} + c_{2} e^{- 4 t} + \frac{t}{2}

(D^{2} + 5 D + 6) y = e^{2 x}

y^{^{''}} - 2 y^{^{'}} + y = e^{x} + e^{2 x}

y^{^{''}} - 2 y^{^{'}} + 5 y = e^{x} (0 = (2 x))

y^{^{''}} - 6 y^{^{'}} + 9 y = 18

y^{^{''}} - y = 8 t e^{t} + 2 e^{t}