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

Lösung

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

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

Schritte zur Lösung

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

Solve linear ODE:

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

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

y^{^{''}} + y = - e^{3} + e^{2 x}

y^{^{''}} - 4 y^{^{'}} + 4 y = 5 t^{2}

y^{^{''}} + 3 y^{^{'}} - 4 y = 3 e^{x}

y^{^{''}} - y^{^{'}} + y = 12 sin (t) - 4 e^{2 t}

y^{^{''}} - y^{^{'}} - 2 y = 1 + t