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受欢迎的微积分 >

inverselaplace 8/((9s^2+9s+4)(s))

解答

的拉普拉斯逆变换

\frac{8}{( 9 s ^{2} + 9 s + 4 ) ( s )}

解答

- 2 e^{- \frac{t}{2}} cos (\frac{7 t}{6}) - \frac{6}{7} e^{- \frac{t}{2}} sin (\frac{7 t}{6}) + 2 H (t)

求解步骤

L^{- 1} {\frac{8}{( s ) ( 9 s ^{2} + 9 s + 4 )}}

将

\frac{8}{( 9 s ^{2} + 9 s + 4 ) s}

用部份分式展开:

\frac{- 18 s - 18}{9 s ^{2} + 9 s + 4} + \frac{2}{s}

= L^{- 1} {\frac{- 18 s - 18}{9 s ^{2} + 9 s + 4} + \frac{2}{s}}

展开

\frac{- 18 s - 18}{9 s ^{2} + 9 s + 4} : - 2 \cdot \frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{7}{36}} - 1 \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{7}{36}}

= L^{- 1} {- 2 \cdot \frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{7}{36}} - 1 \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{7}{36}} + \frac{2}{s}}

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)}

= - 2 L^{- 1} {\frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{7}{36}}} - 1 \cdot L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{7}{36}}} + L^{- 1} {\frac{2}{s}}

L^{- 1} {\frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{7}{36}}} : e^{- \frac{t}{2}} cos (\frac{7 t}{6})

L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{7}{36}}} : e^{- \frac{t}{2}} \frac{6}{7} sin (\frac{7 t}{6})

L^{- 1} {\frac{2}{s}} : 2 H (t)

= - 2 e^{- \frac{t}{2}} cos (\frac{7 t}{6}) - 1 \cdot e^{- \frac{t}{2}} \frac{6}{7} sin (\frac{7 t}{6}) + 2 H (t)

整理

- 2 e^{- \frac{t}{2}} cos (\frac{7 t}{6}) - 1 e^{- \frac{t}{2}} \frac{6}{7} sin (\frac{7 t}{6}) + 2 H (t) : - 2 e^{- \frac{t}{2}} cos (\frac{7 t}{6}) - \frac{6}{7} e^{- \frac{t}{2}} sin (\frac{7 t}{6}) + 2 H (t)

= - 2 e^{- \frac{t}{2}} cos (\frac{7 t}{6}) - \frac{6}{7} e^{- \frac{t}{2}} sin (\frac{7 t}{6}) + 2 H (t)

流行的例子

(\partial)/(\partial x)(sqrt(x))

\frac{\partial}{\partial x} (x)

derivative ((t-4)^3)/6+t^2

d er i v a t i v e \frac{( t - 4 ) ^{3}}{6} + t^{2}

limit as x approaches 0 of 9sin(x)ln(x)

x \to 0 lim (9 sin (x) ln (x))

integral of xln(y)

\int x ln (y) d y

limit as x approaches 1/2 of x/(2x-1)

x \to \frac{1}{2} lim (\frac{x}{2 x - 1})