inverselaplace s/(s^2+4)

Lösung

inverse laplace transformation

\frac{s}{s ^{2} + 4}

cos (2 t)

Schritte zur Lösung

L^{- 1} {\frac{s}{s ^{2} + 4}}

Wende inverse Transformationsregel an: if

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

dann

L^{- 1} {s F (s)} = \frac{d}{d t} f (t) + f (0)

Für

\frac{s}{s ^{2} + 4} : F (s) = \frac{1}{s ^{2} + 4}

= \frac{d}{d t} (L^{- 1} {\frac{1}{s ^{2} + 4}}) + L^{- 1} {\frac{1}{s ^{2} + 4}} (0)

L^{- 1} {\frac{1}{s ^{2} + 4}} : \frac{1}{2} sin (2 t)

\frac{d}{d t} (\frac{1}{2} sin (2 t)) = cos (2 t)

Evaluiere

\frac{1}{2} sin (2 t)

am Punkt

0 = 0

= cos (2 t)

\int \frac{5}{1 + x ^{2}} d x

y^{^{''}} + 6 y^{^{'}} + 5 y = 0

x \to 0 lim (x^{x})

y^{^{''}} - 2 y^{^{'}} + 3 y = 6 sin (3 x) - 6 cos (3 x)

(10 + x^{2}) y^{^{'}} - 2 x y = 0