inverselaplace (e^{-4s})/(s^2+4s+3)

Lösung

inverse laplace transformation

\frac{e ^{- 4 s}}{s ^{2} + 4 s + 3}

H (t - 4) (\frac{1}{2} e^{- t + 4} - \frac{1}{2} e^{- 3 (t - 4)})

Schritte zur Lösung

L^{- 1} {\frac{e ^{- 4 s}}{s ^{2} + 4 s + 3}}

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 - 4) L^{- 1} {\frac{1}{s ^{2} + 4 s + 3}} (t - 4)

L^{- 1} {\frac{1}{s ^{2} + 4 s + 3}} : \frac{1}{2} e^{- t} - \frac{1}{2} e^{- 3 t}

= H (t - 4) (\frac{1}{2} e^{- (t - 4)} - \frac{1}{2} e^{- 3 (t - 4)})

= H (t - 4) (\frac{1}{2} e^{- t + 4} - \frac{1}{2} e^{- 3 (t - 4)})

d er i v a t i v e - 2 x + 3

y^{^{'}} = \frac{x - y}{2}

a re a y = 17 - x^{2}, y = - 2 x + 14

\int \frac{1}{x ^{2} + 2 x - 15} d x

4 y^{^{''}} - 4 y^{^{'}} - 3 y = cos (2 t)