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

Lösung

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

y = c_{1} e^{2 t} + c_{2} e^{- t} + c_{3} e^{t}

Schritte zur Lösung

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- t} + c_{3} e^{t}

y = c_{1} e^{2 t} + c_{2} e^{- t} + c_{3} e^{t}

3 r = \frac{d r}{d θ} - θ^{3}

\int cos (\frac{1}{5} x) d x

\int sin (x) + 1 d x

\frac{d}{d x} (ln (c) os x)

\frac{d x}{d t} = 6 x t^{4}