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

Lösung

inverse laplace transformation

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

e^{2 t} cos (t)

Schritte zur Lösung

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

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

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

= e^{2 t} cos (t)

\frac{\partial}{\partial x} (e^{2 x + y})

t an g e n t f (x) = 7 x - 6 x, at x = 1

x \to 10 lim (12 x)

\frac{d}{d x} (\frac{x}{1 + x ^{\frac{3}{2}}})

\frac{\partial}{\partial x} (\frac{y}{x})