inverselaplace 1/(s^2-2s+3)

Lösung

inverse laplace transformation

\frac{1}{s ^{2} - 2 s + 3}

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

Schritte zur Lösung

L^{- 1} {\frac{1}{s ^{2} - 2 s + 3}}

\frac{1}{s ^{2} - 2 s + 3} = \frac{1}{( s - 1 ) ^{2} + 2}

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

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

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

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

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

\frac{\partial}{\partial x} (3 x^{2} - 2 x + 6 y^{2} + 2 y + 2)

\int (u^{4} - u^{3} + u^{2}) d u

\frac{\partial}{\partial x} (4 y - 2 x y)

d er i v a t i v e sin (2 x)

x \to 2 lim (\frac{1}{( x - 2 )})