inverselaplace (-75s-25)/(1-3s^2)

Lösung

inverse laplace transformation

\frac{- 75 s - 25}{1 - 3 s ^{2}}

\frac{25 3 - 25}{2 3} e^{- \frac{t}{3}} + \frac{25 3 + 25}{2 3} e^{\frac{t}{3}}

Schritte zur Lösung

L^{- 1} {\frac{- 75 s - 25}{1 - 3 s ^{2}}}

Ermittle den Partialbruch von

\frac{- 75 s - 25}{1 - 3 s ^{2}} : \frac{25 3 - 25}{2 ( 3 s + 1 )} + \frac{25 3 + 25}{2 ( 3 s - 1 )}

= L^{- 1} {\frac{25 3 - 25}{2 ( 3 s + 1 )} + \frac{25 3 + 25}{2 ( 3 s - 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{25 3 - 25}{2 ( 3 s + 1 )}} + L^{- 1} {\frac{25 3 + 25}{2 ( 3 s - 1 )}}

L^{- 1} {\frac{25 3 - 25}{2 ( 3 s + 1 )}} : \frac{25 3 - 25}{2 3} e^{- \frac{t}{3}}

L^{- 1} {\frac{25 3 + 25}{2 ( 3 s - 1 )}} : \frac{25 3 + 25}{2 3} e^{\frac{t}{3}}

= \frac{25 3 - 25}{2 3} e^{- \frac{t}{3}} + \frac{25 3 + 25}{2 3} e^{\frac{t}{3}}

\int_{0}^{2} \frac{x}{( 3 - x ) ^{2}} d x

in v erse l a pl a ce \frac{4 s}{s ^{2} + 4}

y^{^{''}} - 3 y^{^{'}} - 2 y = 0

s l o p e x^{5} - x^{3}

y^{^{'}} = \frac{4 x}{y}