laplacetransform t^2sin(1/2 t)

Lösung

laplace transformation

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

- \frac{16 ( 1 - 12 s ^{2} )}{( 4 s ^{2} + 1 ) ^{3}}

Schritte zur Lösung

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

Wende Laplace-Transformationstabelle an:

L {t^{k} f (t)} = (- 1)^{k} \frac{d ^{k}}{d s ^{k}} (L {f (t)})

Für

t^{2} sin (\frac{1}{2} t) : f (t) = sin (\frac{1}{2} t), k = 2

= (- 1)^{2} \frac{d ^{2}}{d s ^{2}} (L {sin (\frac{1}{2} t)})

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

\frac{d ^{2}}{d s ^{2}} (\frac{2}{4 s ^{2} + 1}) = - \frac{16 ( 1 - 12 s ^{2} )}{( 4 s ^{2} + 1 ) ^{3}}

= (- 1)^{2} (- \frac{16 ( 1 - 12 s ^{2} )}{( 4 s ^{2} + 1 ) ^{3}})

= - \frac{16 ( 1 - 12 s ^{2} )}{( 4 s ^{2} + 1 ) ^{3}}

n = 0 \sum \infty n^{n} x^{n}

x \to 0 lim (\frac{3 - 3 cos ( x )}{sin ^{2} ( x )})

d er i v a t i v e f (x) = \frac{11}{( 1 - x ) ^{2}}

\int sin^{3} (2 x) d x

f^{^{''}} (x) = x^{- \frac{3}{2}}