laplacetransform 10cos(t-pi/6)

Lösung

laplace transformation

10 cos (t - \frac{π}{6})

\frac{5 ( 3 s + 1 )}{s ^{2} + 1}

Schritte zur Lösung

L {10 cos (t - \frac{π}{6})}

Verwende die konstante Multiplikationseigenschaft der Laplace-Transformation:

For function

f (t)

and constant

a : L {a \cdot f (t)} = a \cdot L {f (t)}

= 10 L {cos (t - \frac{π}{6})}

L {cos (t - \frac{π}{6})} : \frac{3 s + 1}{2 ( s ^{2} + 1 )}

= 10 \cdot \frac{3 s + 1}{2 ( s ^{2} + 1 )}

Fasse

10 \frac{3 s + 1}{2 ( s ^{2} + 1 )}

zusammen:

\frac{5 ( 3 s + 1 )}{s ^{2} + 1}

= \frac{5 ( 3 s + 1 )}{s ^{2} + 1}

\int_{0}^{π} 13 cos^{6} (x) d x

s l o p e (4, - 3), (- 2, 7)

\int \frac{1}{( 4 x - x ^{2} ) ^{\frac{3}{2}}} d x

l a pl a ce t r an s f or m t e^{- 3 t} sin (5 t)

l a pl a ce t r an s f or m i (t)