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

Lösung

inverse laplace transformation

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

e^{t} cos (t) + e^{t} sin (t)

Schritte zur Lösung

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

Schreibe

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

um:

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

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

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

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

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

= e^{t} cos (t) + 1 \cdot e^{t} sin (t)

Fasse

e^{t} cos (t) + 1 e^{t} sin (t)

zusammen:

e^{t} cos (t) + e^{t} sin (t)

= e^{t} cos (t) + e^{t} sin (t)

\int_{1}^{5} 2 e^{2 x} d x

d er i v a t i v e 3 t + 4 t^{3}

x \to - 0.001 lim (4 x^{2} + 7)

\frac{\partial}{\partial x} (4 x^{3} y^{3} - 18 y - x)

\frac{\partial}{\partial y} (9 x sin (y - z))