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Beliebt Rechnen >

inverselaplace (s+1)/(s^2+25)

Lösung

inverse laplace transformation

\frac{s + 1}{s ^{2} + 25}

Lösung

cos (5 t) + \frac{1}{5} sin (5 t)

Schritte zur Lösung

L^{- 1} {\frac{s + 1}{s ^{2} + 25}}

Schreibe um

= L^{- 1} {\frac{s}{s ^{2} + 25} + \frac{1}{s ^{2} + 25}}

Wende die lineare Eigenschaftder inversen Laplace-Transformation an:

Für Funktionen

f (s), g (s)

und Konstanten

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{s}{s ^{2} + 25}} + L^{- 1} {\frac{1}{s ^{2} + 25}}

L^{- 1} {\frac{s}{s ^{2} + 25}} : cos (5 t)

L^{- 1} {\frac{1}{s ^{2} + 25}} : \frac{1}{5} sin (5 t)

= cos (5 t) + \frac{1}{5} sin (5 t)

Beliebte Beispiele

f(x)=sqrt(1+2e^{3x)}

f (x) = 1 + 2 e^{3 x}

(d^2y)/(dx^2)y+81y=0

\frac{d ^{2} y}{d x ^{2}} y + 81 y = 0

y^{^{'}} = k x

limit as x approaches infinity of 4x+7

x \to \infty lim (4 x + 7)

limit as n approaches infinity of (n^{100})/(50^n)

n \to \infty lim (\frac{n ^{100}}{5 0 ^{n}})