inverselaplace (s-1)/(2s^2+s+3)

解答

的拉普拉斯逆变换

\frac{s - 1}{2 s ^{2} + s + 3}

\frac{1}{2} e^{- \frac{t}{4}} cos (\frac{23 t}{4}) - \frac{5 e ^{- \frac{t}{4}} sin ( \frac{23 t}{4} )}{2 23}

求解步骤

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

展开

\frac{s - 1}{2 s ^{2} + s + 3} : \frac{1}{2} \cdot \frac{s + \frac{1}{4}}{( s + \frac{1}{4} ) ^{2} + \frac{23}{16}} - \frac{5}{8} \cdot \frac{1}{( s + \frac{1}{4} ) ^{2} + \frac{23}{16}}

= L^{- 1} {\frac{1}{2} \cdot \frac{s + \frac{1}{4}}{( s + \frac{1}{4} ) ^{2} + \frac{23}{16}} - \frac{5}{8} \cdot \frac{1}{( s + \frac{1}{4} ) ^{2} + \frac{23}{16}}}

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}{2} L^{- 1} {\frac{s + \frac{1}{4}}{( s + \frac{1}{4} ) ^{2} + \frac{23}{16}}} - \frac{5}{8} L^{- 1} {\frac{1}{( s + \frac{1}{4} ) ^{2} + \frac{23}{16}}}

L^{- 1} {\frac{s + \frac{1}{4}}{( s + \frac{1}{4} ) ^{2} + \frac{23}{16}}} : e^{- \frac{t}{4}} cos (\frac{23 t}{4})

L^{- 1} {\frac{1}{( s + \frac{1}{4} ) ^{2} + \frac{23}{16}}} : e^{- \frac{t}{4}} \frac{4}{23} sin (\frac{23 t}{4})

= \frac{1}{2} e^{- \frac{t}{4}} cos (\frac{23 t}{4}) - \frac{5}{8} e^{- \frac{t}{4}} \frac{4}{23} sin (\frac{23 t}{4})

整理

\frac{1}{2} e^{- \frac{t}{4}} cos (\frac{23 t}{4}) - \frac{5}{8} e^{- \frac{t}{4}} \frac{4}{23} sin (\frac{23 t}{4}) : \frac{1}{2} e^{- \frac{t}{4}} cos (\frac{23 t}{4}) - \frac{5 e ^{- \frac{t}{4}} sin ( \frac{23 t}{4} )}{2 23}

= \frac{1}{2} e^{- \frac{t}{4}} cos (\frac{23 t}{4}) - \frac{5 e ^{- \frac{t}{4}} sin ( \frac{23 t}{4} )}{2 23}

\frac{d}{d x} (5 x e^{x^{2}})

\frac{d}{d t} (2 t e^{2 t})

\int x^{3} 121 - x^{2} d x

t a y l or 6 x e^{x}

d er i v a t i v e 10 (x^{3} - x)^{4}