y^{''}+2y^'+3y=0

Lösung

y^{^{''}} + 2 y^{^{'}} + 3 y = 0

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

Schritte zur Lösung

y^{''} (t) + 2 y^{'} (t) + 3 y = 0

Solve linear ODE:

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

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

a re a y = x^{3} - 3 x^{2} - 27 x + 14, y = x + 14

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