inverselaplace e^{-2s} 1/(s-9)

Lösung

inverse laplace transformation

e^{- 2 s} \frac{1}{s - 9}

H (t - 2) e^{9 (t - 2)}

Schritte zur Lösung

L^{- 1} {e^{- 2 s} \frac{1}{s - 9}}

Wende inverse Transformationsregel an: if

L^{- 1} {F (s)} = f (t)

dann

L^{- 1} {e^{- a s} F (s)} = H (t - a) f (t - a)

H (t)

Heaviside (Schritt) Funktion ist

= H (t - 2) L^{- 1} {\frac{1}{s - 9}} (t - 2)

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

= H (t - 2) e^{9 (t - 2)}

\int \frac{1}{1 + x} d x

s l o p e (0, - 6), (3, - 5)

d er i v a t i v e y = \frac{6}{4 x}

t \to 0 lim (\frac{sin ( 3 t ) + 4 t}{t sec ( t )})

\frac{\partial}{\partial y} (5 x ln (x y))