inverselaplace (3s^2)/((s^2+4)^2)

Lösung

inverse laplace transformation

\frac{3 s ^{2}}{( s ^{2} + 4 ) ^{2}}

\frac{3 sin ( 2 t ) + 6 t cos ( 2 t )}{4}

Schritte zur Lösung

L^{- 1} {\frac{3 s ^{2}}{( s ^{2} + 4 ) ^{2}}}

Ermittle den Partialbruch von

\frac{3 s ^{2}}{( s ^{2} + 4 ) ^{2}} : \frac{3}{s ^{2} + 4} - \frac{12}{( s ^{2} + 4 ) ^{2}}

= L^{- 1} {\frac{3}{s ^{2} + 4} - \frac{12}{( s ^{2} + 4 ) ^{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{3}{s ^{2} + 4}} - L^{- 1} {\frac{12}{( s ^{2} + 4 ) ^{2}}}

L^{- 1} {\frac{3}{s ^{2} + 4}} : \frac{3}{2} sin (2 t)

L^{- 1} {\frac{12}{( s ^{2} + 4 ) ^{2}}} : \frac{3}{4} (sin (2 t) - 2 t cos (2 t))

= \frac{3}{2} sin (2 t) - \frac{3}{4} (sin (2 t) - 2 t cos (2 t))

Vereinfache

\frac{3}{2} sin (2 t) - \frac{3}{4} (sin (2 t) - 2 t cos (2 t)) : \frac{3 sin ( 2 t ) + 6 t cos ( 2 t )}{4}

= \frac{3 sin ( 2 t ) + 6 t cos ( 2 t )}{4}

\int \frac{1}{- x ^{2} + 4 x + 5} d x

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\int \frac{11}{11 + e ^{x}} d x