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^2+8s+16)

Lösung

inverse laplace transformation

\frac{1}{s ^{2} + 8 s + 16}

Lösung

e^{- 4 t} t

Schritte zur Lösung

L^{- 1} {\frac{1}{s ^{2} + 8 s + 16}}

\frac{1}{s ^{2} + 8 s + 16} = \frac{1}{( s + 4 ) ^{2}}

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

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

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

= e^{- 4 t} t

Beliebte Beispiele

derivative of 1/(x+a)

\frac{d}{d x} (\frac{1}{x + a})

integral of (1/(\sqrt[3]{x)})

\int (\frac{1}{3 x}) d x

t a y l or \frac{1}{x - 1}

integral of ln(e^{2x-1})

\int ln (e^{2 x - 1}) d x

(\partial)/(\partial x)(2x+2y^2)

\frac{\partial}{\partial x} (2 x + 2 y^{2})