inverselaplace ((s-10))/((s^2-2s-3))

Lösung

inverse laplace transformation

\frac{( s - 10 )}{( s ^{2} - 2 s - 3 )}

\frac{11}{4} e^{- t} - \frac{7}{4} e^{3 t}

Schritte zur Lösung

L^{- 1} {\frac{( s - 10 )}{( s ^{2} - 2 s - 3 )}}

Ermittle den Partialbruch von

\frac{s - 10}{s ^{2} - 2 s - 3} : \frac{11}{4 ( s + 1 )} - \frac{7}{4 ( s - 3 )}

= L^{- 1} {\frac{11}{4 ( s + 1 )} - \frac{7}{4 ( s - 3 )}}

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{11}{4 ( s + 1 )}} - L^{- 1} {\frac{7}{4 ( s - 3 )}}

L^{- 1} {\frac{11}{4 ( s + 1 )}} : \frac{11}{4} e^{- t}

L^{- 1} {\frac{7}{4 ( s - 3 )}} : \frac{7}{4} e^{3 t}

= \frac{11}{4} e^{- t} - \frac{7}{4} e^{3 t}

im pl i c i t \frac{d y}{d x}, x^{6} - 5 x y^{3} = 9 x y

x \to 0 lim (arctan (\frac{1}{x}))

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

a re a \frac{1}{( 1 - 7 x )}, - 6 < x < 0

\frac{\partial}{\partial x} (4 x^{- x^{2} - y^{2} - z^{2}})