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Beliebt Rechnen >

inverselaplace 1/(s^2(s^2+5))

Lösung

inverse laplace transformation

\frac{1}{s ^{2} ( s ^{2} + 5 )}

Lösung

\frac{t}{5} - \frac{1}{5 5} sin (5 t)

Schritte zur Lösung

L^{- 1} {\frac{1}{s ^{2} ( s ^{2} + 5 )}}

Ermittle den Partialbruch von

\frac{1}{s ^{2} ( s ^{2} + 5 )} : \frac{1}{5 s ^{2}} - \frac{1}{5 ( s ^{2} + 5 )}

= L^{- 1} {\frac{1}{5 s ^{2}} - \frac{1}{5 ( s ^{2} + 5 )}}

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{1}{5 s ^{2}}} - L^{- 1} {\frac{1}{5 ( s ^{2} + 5 )}}

L^{- 1} {\frac{1}{5 s ^{2}}} : \frac{t}{5}

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

= \frac{t}{5} - \frac{1}{5 5} sin (5 t)

Beliebte Beispiele

integral of cos(1-2x)

\int cos (1 - 2 x) d x

tangent 4q+1,\at x=10

t an g e n t 4 q + 1, at x = 10

5x^2(d^2y)/(dx^2)-16x(dy)/(dx)+4y=0

5 x^{2} \frac{d ^{2} y}{d x ^{2}} - 16 x \frac{d y}{d x} + 4 y = 0

integral from 0 to 5 of xsqrt(5-x)

\int_{0}^{5} x 5 - x d x

x^2y^{''}+2xy^'-6y=0,y(1)=0

x^{2} y^{^{''}} + 2 x y^{^{'}} - 6 y = 0, y (1) = 0