inverselaplace 2/((s-4)^2)

Lösung

inverse laplace transformation

\frac{2}{( s - 4 ) ^{2}}

e^{4 t} \cdot 2 t

Schritte zur Lösung

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

Wende inverse Transformationsregel an: if

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

dann

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

Für

\frac{2}{( s - 4 ) ^{2}} : a = 4, F (s) = \frac{2}{s ^{2}}

= e^{4 t} L^{- 1} {\frac{2}{s ^{2}}}

L^{- 1} {\frac{2}{s ^{2}}} : 2 t

= e^{4 t} \cdot 2 t

\frac{d}{d x} (\frac{1}{1 + e ^{\frac{1}{x}}})

x \to 3 - lim (\frac{∣ x ∣}{x})

- x y + (3 + x) (\frac{d y}{d x}) = 0

n = 1 \sum \infty 8 (0.8)^{n - 1}

\int \frac{5 x ^{2} - 3 x + 2}{x ^{3} - 2 x ^{2}} d x