inverselaplace (5s-2)/(7s^2-s-11)

解答

的拉普拉斯逆变换

\frac{5 s - 2}{7 s ^{2} - s - 11}

\frac{5}{7} e^{\frac{t}{14}} cosh (\frac{309 t}{14}) - \frac{23 e ^{\frac{t}{14}} sinh ( \frac{309 t}{14} )}{7 309}

求解步骤

L^{- 1} {\frac{5 s - 2}{7 s ^{2} - s - 11}}

展开

\frac{5 s - 2}{7 s ^{2} - s - 11} : \frac{5}{7} \cdot \frac{s - \frac{1}{14}}{( s - \frac{1}{14} ) ^{2} - \frac{309}{196}} - \frac{23}{98} \cdot \frac{1}{( s - \frac{1}{14} ) ^{2} - \frac{309}{196}}

= L^{- 1} {\frac{5}{7} \cdot \frac{s - \frac{1}{14}}{( s - \frac{1}{14} ) ^{2} - \frac{309}{196}} - \frac{23}{98} \cdot \frac{1}{( s - \frac{1}{14} ) ^{2} - \frac{309}{196}}}

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{5}{7} L^{- 1} {\frac{s - \frac{1}{14}}{( s - \frac{1}{14} ) ^{2} - \frac{309}{196}}} - \frac{23}{98} L^{- 1} {\frac{1}{( s - \frac{1}{14} ) ^{2} - \frac{309}{196}}}

L^{- 1} {\frac{s - \frac{1}{14}}{( s - \frac{1}{14} ) ^{2} - \frac{309}{196}}} : e^{\frac{t}{14}} cosh (\frac{309 t}{14})

L^{- 1} {\frac{1}{( s - \frac{1}{14} ) ^{2} - \frac{309}{196}}} : e^{\frac{t}{14}} \frac{14}{309} sinh (\frac{309 t}{14})

= \frac{5}{7} e^{\frac{t}{14}} cosh (\frac{309 t}{14}) - \frac{23}{98} e^{\frac{t}{14}} \frac{14}{309} sinh (\frac{309 t}{14})

整理

\frac{5}{7} e^{\frac{t}{14}} cosh (\frac{309 t}{14}) - \frac{23}{98} e^{\frac{t}{14}} \frac{14}{309} sinh (\frac{309 t}{14}) : \frac{5}{7} e^{\frac{t}{14}} cosh (\frac{309 t}{14}) - \frac{23 e ^{\frac{t}{14}} sinh ( \frac{309 t}{14} )}{7 309}

= \frac{5}{7} e^{\frac{t}{14}} cosh (\frac{309 t}{14}) - \frac{23 e ^{\frac{t}{14}} sinh ( \frac{309 t}{14} )}{7 309}

x \to - \infty lim (x - 2)

\int (x - 6)^{2} (x + 8) d x

\frac{\partial}{\partial x} (y - ln (x y))

\frac{d ^{2}}{d x ^{2}} ((14 t^{2} + 6)^{0.5})

10 x^{^{''}} + 1000 x = 0