inverselaplace x/((x-1)^2)

Lösung

inverse laplace transformation

\frac{x}{( x - 1 ) ^{2}}

e^{t} + e^{t} t

Schritte zur Lösung

L^{- 1} {\frac{x}{( x - 1 ) ^{2}}}

Ermittle den Partialbruch von

\frac{x}{( x - 1 ) ^{2}} : \frac{1}{x - 1} + \frac{1}{( x - 1 ) ^{2}}

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

Wende die lineare Eigenschaftder inversen Laplace-Transformation an:

Für Funktionen

f (s), g (s)

und Konstanten

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

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

L^{- 1} {\frac{1}{x - 1}} : e^{t}

L^{- 1} {\frac{1}{( x - 1 ) ^{2}}} : e^{t} t

= e^{t} + e^{t} t

n = 0 \sum \infty p^{k}

n = 1 \sum \infty \frac{n - 8}{n + 8}

\int (e^{x} + a^{x}) d x

\int \frac{1}{( 1 + x )} d x

\int_{0}^{π} sin (2 x) cos (x) d x