inverselaplace (s-4)/((s^2+5)^2)

Lösung

inverse laplace transformation

\frac{s - 4}{( s ^{2} + 5 ) ^{2}}

\frac{t sin ( 5 t )}{2 5} - \frac{2}{5 5} (sin (5 t) - 5 t cos (5 t))

Schritte zur Lösung

L^{- 1} {\frac{s - 4}{( s ^{2} + 5 ) ^{2}}}

Schreibe um

= L^{- 1} {\frac{s}{( s ^{2} + 5 ) ^{2}} - \frac{4}{( s ^{2} + 5 ) ^{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{s}{( s ^{2} + 5 ) ^{2}}} - 4 L^{- 1} {\frac{1}{( s ^{2} + 5 ) ^{2}}}

L^{- 1} {\frac{s}{( s ^{2} + 5 ) ^{2}}} : \frac{t sin ( 5 t )}{2 5}

L^{- 1} {\frac{1}{( s ^{2} + 5 ) ^{2}}} : \frac{1}{10 5} (sin (5 t) - 5 t cos (5 t))

= \frac{t sin ( 5 t )}{2 5} - 4 \cdot \frac{1}{10 5} (sin (5 t) - 5 t cos (5 t))

Fasse

\frac{t sin ( 5 t )}{2 5} - 4 \frac{1}{10 5} (sin (5 t) - 5 t cos (5 t))

zusammen:

\frac{t sin ( 5 t )}{2 5} - \frac{2}{5 5} (sin (5 t) - 5 t cos (5 t))

= \frac{t sin ( 5 t )}{2 5} - \frac{2}{5 5} (sin (5 t) - 5 t cos (5 t))

\int_{0}^{4} x x^{2} + 9 d x

\int \frac{1}{x ^{4} x ^{2} + 3} d x

x \to 0 lim (\frac{x ^{3}}{x ^{4}})

n or ma l y = 5 x^{2}, (- 2, 20)

\frac{d}{d x} (f (x) (x^{2}))