laplacetransform sin(1/2 t)*t^2

Lösung

laplace transformation

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

- \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

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

\int - cos (x) 2 x d x

y^{^{'}} = 4 y

\frac{d}{d x} (A cos (2 x))

\int sin (nπ x) d x

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