laplacetransform tsin(3t)

解答

拉普拉斯变换

t sin (3 t)

\frac{6 s}{( s ^{2} + 9 ) ^{2}}

求解步骤

L {t sin (3 t)}

使用拉普拉斯变换表:

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

对于

t sin (3 t) : f (t) = sin (3 t), k = 1

= (- 1)^{1} \frac{d}{d s} (L {sin (3 t)})

L {sin (3 t)} : \frac{3}{s ^{2} + 9}

\frac{d}{d s} (\frac{3}{s ^{2} + 9}) = - \frac{6 s}{( s ^{2} + 9 ) ^{2}}

= (- 1)^{1} (- \frac{6 s}{( s ^{2} + 9 ) ^{2}})

= \frac{6 s}{( s ^{2} + 9 ) ^{2}}

\int (x^{10} - 5 x^{2})^{3} (x^{9} - x) d x

x y^{^{'}} + 2 y - 12 = 0

\frac{d}{d x} (ln (\frac{x ^{3}}{e ^{3 x}}))

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

f (θ) = sec (θ)