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受欢迎的微积分 >

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

解答

的拉普拉斯逆变换

\frac{( s + 1 )}{s ( s ^{2} + s + 1 )}

解答

H (t) - e^{- \frac{t}{2}} cos (\frac{3 t}{2}) - \frac{sin ( \frac{3 t}{2} )}{3}

求解步骤

L^{- 1} {\frac{( s + 1 )}{s ( s ^{2} + s + 1 )}}

将

\frac{s + 1}{s ( s ^{2} + s + 1 )}

用部份分式展开:

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

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

展开

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

= L^{- 1} {\frac{1}{s} - (\frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{3}{4}} - \frac{1}{2} \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{3}{4}})}

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)}

= L^{- 1} {\frac{1}{s}} - L^{- 1} {\frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{3}{4}} - \frac{1}{2} \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{3}{4}}}

L^{- 1} {\frac{1}{s}} : H (t)

L^{- 1} {\frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{3}{4}} - \frac{1}{2} \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{3}{4}}} : e^{- \frac{t}{2}} cos (\frac{3 t}{2}) - \frac{sin ( \frac{3 t}{2} )}{3}

= H (t) - e^{- \frac{t}{2}} cos (\frac{3 t}{2}) - \frac{sin ( \frac{3 t}{2} )}{3}

流行的例子

\frac{d y}{d x} + 1 = 2 y

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\int (15 x^{- 1} - 4 sinh (x)) d x

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\frac{d}{d x} ((x^{6} + 1)^{\frac{5}{2}})

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\int tan^{3} (x) sec^{4} (x) d x