laplacetransform t^2cos(2t+30)

Lösung

laplace transformation

t^{2} cos (2 t + 3 0^{\circ})

\frac{3 s ^{3} - 6 s ^{2} - 12 3 s + 8}{( s ^{2} + 4 ) ^{3}}

Schritte zur Lösung

L {t^{2} cos (2 t + 3 0^{\circ})}

Wende Laplace-Transformationstabelle an:

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

Für

t^{2} cos (2 t + 3 0^{\circ}) : f (t) = cos (2 t + 3 0^{\circ}), k = 2

= (- 1)^{2} \frac{d ^{2}}{d s ^{2}} (L {cos (2 t + 3 0^{\circ})})

L {cos (2 t + 3 0^{\circ})} : \frac{3 s - 2}{2 ( s ^{2} + 4 )}

\frac{d ^{2}}{d s ^{2}} (\frac{3 s - 2}{2 ( s ^{2} + 4 )}) = \frac{3 s ^{3} - 6 s ^{2} - 12 3 s + 8}{( s ^{2} + 4 ) ^{3}}

= (- 1)^{2} \frac{3 s ^{3} - 6 s ^{2} - 12 3 s + 8}{( s ^{2} + 4 ) ^{3}}

= \frac{3 s ^{3} - 6 s ^{2} - 12 3 s + 8}{( s ^{2} + 4 ) ^{3}}

d er i v a t i v e f (x) = 5 x sin (π x)

y^{^{''}} - 2 y^{^{'}} + y = e^{x} tan^{- 1} (x)

x \to - 2 lim ([f (x) + g (x)])

l a pl a ce t r an s f or m - 5

\int_{0}^{2 π} cos^{3} (θ) d θ