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受欢迎的微积分 >

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

解答

的拉普拉斯逆变换

\frac{( s ^{2} )}{s ^{3} - 1}

解答

\frac{1}{3} e^{t} + \frac{2}{3} e^{- \frac{t}{2}} cos (\frac{3 t}{2})

求解步骤

L^{- 1} {\frac{( s ^{2} )}{s ^{3} - 1}}

将

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

用部份分式展开:

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

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

展开

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

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

L^{- 1} {\frac{1}{3 ( s - 1 )}} : \frac{1}{3} e^{t}

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

= \frac{1}{3} e^{t} + \frac{1}{3} e^{- \frac{t}{2}} \cdot 2 cos (\frac{3 t}{2})

整理

\frac{1}{3} e^{t} + \frac{1}{3} e^{- \frac{t}{2}} 2 cos (\frac{3 t}{2}) : \frac{1}{3} e^{t} + \frac{2}{3} e^{- \frac{t}{2}} cos (\frac{3 t}{2})

= \frac{1}{3} e^{t} + \frac{2}{3} e^{- \frac{t}{2}} cos (\frac{3 t}{2})

流行的例子

(\partial)/(\partial x)(e^{-(x^2+y^2)})

\frac{\partial}{\partial x} (e^{- (x^{2} + y^{2})})

limit as x approaches 8 of x-2

x \to 8 lim (x - 2)

limit as x approaches 2 of 7/((x-2)^2)

x \to 2 lim (\frac{7}{( x - 2 ) ^{2}})

cos(x)y^'+sin(x)y=2cos^3(x)sin(x)-1

cos (x) y^{^{'}} + sin (x) y = 2 cos^{3} (x) sin (x) - 1

limit as x approaches infinity of 3

x \to \infty lim (3)