inverselaplace (-3)/((s-3)^2+4)

Lösung

inverse laplace transformation

\frac{- 3}{( s - 3 ) ^{2} + 4}

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

Schritte zur Lösung

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

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

L^{- 1} {\frac{- 3}{s ^{2} + 4}} : - \frac{3}{2} sin (2 t)

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

\int_{0}^{\infty} \frac{1}{1 + x ^{2}} d x

x \to 0 lim (\frac{4 x}{tan ( x )})

\int [\frac{100}{25 + ( 5 y ) ^{2}} + e^{4}] d y

y^{^{'}} = e^{2 x} + 3 y

\int \frac{1}{x ^{2} + a ^{2}} d x