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

Lösung

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

y = c_{1} e^{t} + c_{2} t e^{t} + c_{3} e^{- t} + c_{4} t e^{- t}

Schritte zur Lösung

y^{(4)} (t) - 2 y^{''} (t) + y = 0

Solve linear ODE:

y = c_{1} e^{t} + c_{2} t e^{t} + c_{3} e^{- t} + c_{4} t e^{- t}

y = c_{1} e^{t} + c_{2} t e^{t} + c_{3} e^{- t} + c_{4} t e^{- t}

\frac{d}{d x} (3 cos^{2} (x) - 6 sin (x))

\frac{\partial}{\partial x} (arcsin (3 x))

\frac{d ^{2}}{d x ^{2}} (\frac{x}{x - 1})

x \to 0 + lim (3 x)

a re a e^{x} - 2, 0 \leq x \leq 2