inverselaplace 1/(s^2-3s+5)

Lösung

inverse laplace transformation

\frac{1}{s ^{2} - 3 s + 5}

e^{\frac{3 t}{2}} \frac{2}{11} sin (\frac{11 t}{2})

Schritte zur Lösung

L^{- 1} {\frac{1}{s ^{2} - 3 s + 5}}

\frac{1}{s ^{2} - 3 s + 5} = \frac{1}{( s - \frac{3}{2} ) ^{2} + \frac{11}{4}}

= L^{- 1} {\frac{1}{( s - \frac{3}{2} ) ^{2} + \frac{11}{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 - \frac{3}{2} ) ^{2} + \frac{11}{4}} : a = \frac{3}{2}, F (s) = \frac{1}{s ^{2} + \frac{11}{4}}

= e^{\frac{3}{2} t} L^{- 1} {\frac{1}{s ^{2} + \frac{11}{4}}}

L^{- 1} {\frac{1}{s ^{2} + \frac{11}{4}}} : \frac{2}{11} sin (\frac{11 t}{2})

= e^{\frac{3 t}{2}} \frac{2}{11} sin (\frac{11 t}{2})

y^{^{'}} = \frac{( 4 t ^{3} + 1 )}{( 2 y - 6 )}

x \to - 1 lim (\frac{2 x}{∣ x ∣})

\int \frac{e ^{x^{\frac{1}{2}}}}{x ^{3}} d x

\int (- 2 T + x) \cdot e^{- i \cdot z \cdot x} d x

l a pl a ce t r an s f or m (t^{2}) \cdot (sin (2 \cdot t))