laplacetransform e^{-t}t^2

Lösung

laplace transformation

e^{- t} t^{2}

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

Schritte zur Lösung

L {e^{- t} t^{2}}

Wende Laplace-Transformationstabelle an:

L {t^{k} f (t)} = (- 1)^{k} \frac{d ^{k}}{d s ^{k}} (L {f (t)})

Für

e^{- t} t^{2} : f (t) = e^{- t}, k = 2

= (- 1)^{2} \frac{d ^{2}}{d s ^{2}} (L {e^{- t}})

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

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

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

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

\int 2 x 2 - x^{2} d x

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

n = 1 \sum \infty \frac{n}{4 n ^{3} + 1}

\int \frac{e ^{x}}{1 - 25 e ^{2 x}} d x

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