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

Lösung

inverse laplace transformation

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

e^{t} \frac{1}{2} sin (2 t)

Schritte zur Lösung

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

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

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

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 - 1 ) ^{2} + 4} : a = 1, F (s) = \frac{1}{s ^{2} + 4}

= e^{t} L^{- 1} {\frac{1}{s ^{2} + 4}}

L^{- 1} {\frac{1}{s ^{2} + 4}} : \frac{1}{2} sin (2 t)

= e^{t} \frac{1}{2} sin (2 t)

t an g e n t f (x) = 25 - x, at x = 0

\int (3^{- x}) d x

im pl i c i t \frac{d y}{d x}, y = x^{2} + 3

\frac{d}{d x} (x sin (\frac{π}{x}))

f (x) = tan (ln (x))