de

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Beliebt Rechnen >

inverselaplace 1/((s-1)^2+4)

Lösung

inverse laplace transformation

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

Lösung

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

Schritte zur Lösung

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

Wende inverse Transformationsregel an: if

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

dann

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

Für

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

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

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

= e^{t} \frac{1}{2} sin (2 t)

Beliebte Beispiele

tangent f(x)=x^2+1,\at x=-5

t an g e n t f (x) = x^{2} + 1, at x = - 5

integral from 0 to 1 of cos(pix)

\int_{0}^{1} cos (π x) d x

limit as x approaches 3 of 2x^3-4x-2

x \to 3 lim (2 x^{3} - 4 x - 2)

integral of x^{35}sin(x^{36})

\int x^{35} sin (x^{36}) d x

laplacetransform-4cos(5t)

l a pl a ce t r an s f or m - 4 cos (5 t)