inverselaplace 1/(s^2-4s+5)

Lösung

inverse laplace transformation

\frac{1}{s ^{2} - 4 s + 5}

e^{2 t} sin (t)

Schritte zur Lösung

L^{- 1} {\frac{1}{s ^{2} - 4 s + 5}}

\frac{1}{s ^{2} - 4 s + 5} = \frac{1}{( s - 2 ) ^{2} + 1}

= L^{- 1} {\frac{1}{( s - 2 ) ^{2} + 1}}

Wende inverse Transformationsregel an: if

L^{- 1} {F (s)} = f (t)

dann

L^{- 1} {F (s - a)} = e^{a t} f (t)

Für

\frac{1}{( s - 2 ) ^{2} + 1} : a = 2, F (s) = \frac{1}{s ^{2} + 1}

= e^{2 t} L^{- 1} {\frac{1}{s ^{2} + 1}}

L^{- 1} {\frac{1}{s ^{2} + 1}} : sin (t)

= e^{2 t} sin (t)

x \to 2 lim (\frac{2 x}{x - 2})

\int x^{2} 6 x + 4 d x

d er i v a t i v e - \frac{3 x}{2}

\frac{d}{d x} (2 x z (x) cos (sin (x)))

\frac{d x}{d t} = 3 (x^{2} + 1), x (\frac{π}{4}) = 1