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

Lösung

inverse laplace transformation

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

\frac{1}{2} (sin (t) - t cos (t))

Schritte zur Lösung

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

= L^{- 1} {\frac{1}{2} \cdot \frac{2 \cdot 1 ^{3}}{( x ^{2} + 1 ^{2} ) ^{2}}}

Wende die konstante Multiplikationseigenschaft der inversen Laplace-Transformation an:

Für Funktionen

f (t)

und Konstante

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= \frac{1}{2} L^{- 1} {\frac{2 \cdot 1 ^{3}}{( x ^{2} + 1 ^{2} ) ^{2}}}

Verwende inverse Laplace-Transformationstabelle:

L^{- 1} {\frac{2 a ^{3}}{( s ^{2} + a ^{2} ) ^{2}}} = sin (a t) - a t cos (a t)

L^{- 1} {\frac{2 \cdot 1 ^{3}}{( x ^{2} + 1 ^{2} ) ^{2}}} = sin (t) - t cos (t)

= \frac{1}{2} (sin (t) - t cos (t))

\frac{\partial}{\partial y} (e^{x - 2 y})

\int_{0}^{1} x y d x

y^{^{''}} - (\frac{3}{x}) y^{^{'}} - (\frac{6}{x ^{2}}) y = 0

im pl i c i t ln (5 + e^{x y}) = y

\frac{\partial}{\partial x} (ln (2 x + 2 c t))