inverselaplace (s-6)/(3s^2+s+12)

解答

的拉普拉斯逆变换

\frac{s - 6}{3 s ^{2} + s + 12}

\frac{1}{3} e^{- \frac{t}{6}} cos (\frac{143 t}{6}) - \frac{37 e ^{- \frac{t}{6}} sin ( \frac{143 t}{6} )}{3 143}

求解步骤

L^{- 1} {\frac{s - 6}{3 s ^{2} + s + 12}}

展开

\frac{s - 6}{3 s ^{2} + s + 12} : \frac{1}{3} \cdot \frac{s + \frac{1}{6}}{( s + \frac{1}{6} ) ^{2} + \frac{143}{36}} - \frac{37}{18} \cdot \frac{1}{( s + \frac{1}{6} ) ^{2} + \frac{143}{36}}

= L^{- 1} {\frac{1}{3} \cdot \frac{s + \frac{1}{6}}{( s + \frac{1}{6} ) ^{2} + \frac{143}{36}} - \frac{37}{18} \cdot \frac{1}{( s + \frac{1}{6} ) ^{2} + \frac{143}{36}}}

利用逆拉普拉斯变换的线性特性:

对于函数

f (s), g (s)

和常数

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= \frac{1}{3} L^{- 1} {\frac{s + \frac{1}{6}}{( s + \frac{1}{6} ) ^{2} + \frac{143}{36}}} - \frac{37}{18} L^{- 1} {\frac{1}{( s + \frac{1}{6} ) ^{2} + \frac{143}{36}}}

L^{- 1} {\frac{s + \frac{1}{6}}{( s + \frac{1}{6} ) ^{2} + \frac{143}{36}}} : e^{- \frac{t}{6}} cos (\frac{143 t}{6})

L^{- 1} {\frac{1}{( s + \frac{1}{6} ) ^{2} + \frac{143}{36}}} : e^{- \frac{t}{6}} \frac{6}{143} sin (\frac{143 t}{6})

= \frac{1}{3} e^{- \frac{t}{6}} cos (\frac{143 t}{6}) - \frac{37}{18} e^{- \frac{t}{6}} \frac{6}{143} sin (\frac{143 t}{6})

整理

\frac{1}{3} e^{- \frac{t}{6}} cos (\frac{143 t}{6}) - \frac{37}{18} e^{- \frac{t}{6}} \frac{6}{143} sin (\frac{143 t}{6}) : \frac{1}{3} e^{- \frac{t}{6}} cos (\frac{143 t}{6}) - \frac{37 e ^{- \frac{t}{6}} sin ( \frac{143 t}{6} )}{3 143}

= \frac{1}{3} e^{- \frac{t}{6}} cos (\frac{143 t}{6}) - \frac{37 e ^{- \frac{t}{6}} sin ( \frac{143 t}{6} )}{3 143}

\int e^{\frac{x}{2} + 3} d x

\int \frac{2 y - 1}{y} d y

d er i v a t i v e f (x) = e^{5}

\frac{d}{d x} (- \frac{128 x}{( 16 x ^{2} + 1 ) ^{2}})

f (t) = \frac{2 t}{1 + t ^{2}}