laplacetransform (t-(3pi)/6)sin(6t)

解答

拉普拉斯变换

(t - \frac{3 π}{6}) sin (6 t)

\frac{12 s}{( s ^{2} + 36 ) ^{2}} - \frac{3 π}{s ^{2} + 36}

求解步骤

L {sin (6 t) (t - \frac{3 π}{6})}

展开

(t - \frac{3 π}{6}) sin (6 t) : t sin (6 t) - \frac{π}{2} sin (6 t)

= L {t sin (6 t) - \frac{π}{2} sin (6 t)}

利用拉普拉斯变换的线性特性:

对于函数

f (t), g (t)

和常数

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= L {t sin (6 t)} - \frac{π}{2} L {sin (6 t)}

L {t sin (6 t)} : \frac{12 s}{( s ^{2} + 36 ) ^{2}}

L {sin (6 t)} : \frac{6}{s ^{2} + 36}

= \frac{12 s}{( s ^{2} + 36 ) ^{2}} - \frac{π}{2} \cdot \frac{6}{s ^{2} + 36}

整理

\frac{12 s}{( s ^{2} + 36 ) ^{2}} - \frac{π}{2} \frac{6}{s ^{2} + 36} : \frac{12 s}{( s ^{2} + 36 ) ^{2}} - \frac{3 π}{s ^{2} + 36}

= \frac{12 s}{( s ^{2} + 36 ) ^{2}} - \frac{3 π}{s ^{2} + 36}

\frac{d}{d x} (6 ln (cos (x)))

\frac{\partial}{\partial x} (e^{x y} sin (y))

x \to 0 lim (9 x^{2} - 2 x + 5)

\frac{\partial}{\partial b} (1 0^{- 2 b})

\int \frac{x ^{2}}{x ^{2} + x - 6} d x