inverselaplace (e^{-2x})/(2x^2+1)

Lösung

inverse laplace transformation

\frac{e ^{- 2 x}}{2 x ^{2} + 1}

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

Schritte zur Lösung

L^{- 1} {\frac{e ^{- 2 x}}{2 x ^{2} + 1}}

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}{2 x ^{2} + 1}} (t - 2)

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

= H (t - 2) \frac{1}{2} sin (\frac{1}{2} (t - 2))

y^{^{''}} - 3 y^{^{'}} - 10 y = 4 e^{4 t}, y (0) = 0, y^{^{'}} (0) = 0

n = 9 \sum \infty e^{5 - 6 n}

\int (\frac{1}{1 + 9 x ^{2}}) d x

\frac{d}{d x} (5 sin (- 4 x))

\int \frac{1}{25 - 36 x ^{2}} d x