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Beliebt Rechnen >

inverselaplace 1/((s-3)(s-9))

Lösung

inverse laplace transformation

\frac{1}{( s - 3 ) ( s - 9 )}

Lösung

- \frac{1}{6} e^{3 t} + \frac{1}{6} e^{9 t}

Schritte zur Lösung

L^{- 1} {\frac{1}{( s - 3 ) ( s - 9 )}}

Ermittle den Partialbruch von

\frac{1}{( s - 3 ) ( s - 9 )} : - \frac{1}{6 ( s - 3 )} + \frac{1}{6 ( s - 9 )}

= L^{- 1} {- \frac{1}{6 ( s - 3 )} + \frac{1}{6 ( s - 9 )}}

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{1}{6 ( s - 3 )}} + L^{- 1} {\frac{1}{6 ( s - 9 )}}

L^{- 1} {\frac{1}{6 ( s - 3 )}} : \frac{1}{6} e^{3 t}

L^{- 1} {\frac{1}{6 ( s - 9 )}} : \frac{1}{6} e^{9 t}

= - \frac{1}{6} e^{3 t} + \frac{1}{6} e^{9 t}

Beliebte Beispiele

limit as (x,y) approaches (0,0) of ((sin(2x)-2x+y))/(x^3+y)

(x, y) \to (0, 0) lim (\frac{( sin ( 2 x ) - 2 x + y )}{x ^{3} + y})

integral of ln(x)*ln(x)

\int ln (x) \cdot ln (x) d x

y^{''}x=-5sin(8x)

y^{^{''}} x = - 5 sin (8 x)

(\partial)/(\partial x)(xln(z^7+x^5))

\frac{\partial}{\partial x} (x ln (z^{7} + x^{5}))

derivative f(x)=(x^2+4x+3)^2

d er i v a t i v e f (x) = (x^{2} + 4 x + 3)^{2}