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Beliebt Rechnen >

inverselaplace s/((s^2+16)^2)

Lösung

inverse laplace transformation

\frac{s}{( s ^{2} + 16 ) ^{2}}

Lösung

\frac{t sin ( 4 t )}{8}

Schritte zur Lösung

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

Wende inverse Transformationsregel an: if

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

dann

L^{- 1} {s F (s)} = \frac{d}{d t} f (t) + f (0)

Für

\frac{s}{( s ^{2} + 16 ) ^{2}} : F (s) = \frac{1}{( s ^{2} + 16 ) ^{2}}

= \frac{d}{d t} (L^{- 1} {\frac{1}{( s ^{2} + 16 ) ^{2}}}) + L^{- 1} {\frac{1}{( s ^{2} + 16 ) ^{2}}} (0)

L^{- 1} {\frac{1}{( s ^{2} + 16 ) ^{2}}} : \frac{1}{128} (sin (4 t) - 4 t cos (4 t))

\frac{d}{d t} (\frac{1}{128} (sin (4 t) - 4 t cos (4 t))) = \frac{t sin ( 4 t )}{8}

Evaluiere

\frac{1}{128} (sin (4 t) - 4 t cos (4 t))

am Punkt

0 = 0

= \frac{t sin ( 4 t )}{8}

Beliebte Beispiele

4y^{''}+6y^'+7y=0

4 y^{^{''}} + 6 y^{^{'}} + 7 y = 0

derivative of sin(e^{2xy}+e^{sin(2xy)})

\frac{d}{d x} (sin (e^{2 x y}) + e^{s i n (2 x y)})

tangent y=sqrt(4x+33),(4,7)

t an g e n t y = 4 x + 33, (4, 7)

integral of (2x)^{1/2}

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

integral of x^2sqrt(1+(2x)^2)

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