laplacetransform \H(t-2pi)sin(t-2pi)

解答

拉普拉斯变换

H (t - 2 π) sin (t - 2 π)

\frac{e ^{- 2 π s}}{s ^{2} + 1}

求解步骤

L {H (t - 2 π) sin (t - 2 π)}

使用拉普拉斯变换表:

L {H (t - c) f (t)} = e^{- cs} L {f (t + c)}

对于

H (t - 2 π) sin (t - 2 π) : f (t) = sin (t - 2 π), c = 2 π

= e^{- 2 π s} L {sin ((t + 2 π) - 2 π)}

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

= e^{- 2 π s} \frac{1}{s ^{2} + 1}

= \frac{e ^{- 2 π s}}{s ^{2} + 1}

\frac{d}{d x} (\frac{2 + x}{( 2 - x ) ^{5}})

\int_{1}^{5} \frac{M}{e ^{M}} d M

\int \frac{1}{x ( a + b x )} d x

x \to 1 lim (\frac{x - 8}{3 x - 4})

\int (x^{2} - 5 x) e^{x} d x