inverselaplace (9s)/(5(s^2+5s+11))

解答

的拉普拉斯逆变换

\frac{9 s}{5 ( s ^{2} + 5 s + 11 )}

\frac{9}{5} e^{- \frac{5 t}{2}} cos (\frac{19 t}{2}) - \frac{9 e ^{- \frac{5 t}{2}} sin ( \frac{19 t}{2} )}{19}

求解步骤

L^{- 1} {\frac{9 s}{5 ( s ^{2} + 5 s + 11 )}}

展开

\frac{9 s}{5 ( s ^{2} + 5 s + 11 )} : \frac{9}{5} \cdot \frac{s + \frac{5}{2}}{( s + \frac{5}{2} ) ^{2} + \frac{19}{4}} - \frac{9}{2} \cdot \frac{1}{( s + \frac{5}{2} ) ^{2} + \frac{19}{4}}

= L^{- 1} {\frac{9}{5} \cdot \frac{s + \frac{5}{2}}{( s + \frac{5}{2} ) ^{2} + \frac{19}{4}} - \frac{9}{2} \cdot \frac{1}{( s + \frac{5}{2} ) ^{2} + \frac{19}{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)}

= \frac{9}{5} L^{- 1} {\frac{s + \frac{5}{2}}{( s + \frac{5}{2} ) ^{2} + \frac{19}{4}}} - \frac{9}{2} L^{- 1} {\frac{1}{( s + \frac{5}{2} ) ^{2} + \frac{19}{4}}}

L^{- 1} {\frac{s + \frac{5}{2}}{( s + \frac{5}{2} ) ^{2} + \frac{19}{4}}} : e^{- \frac{5 t}{2}} cos (\frac{19 t}{2})

L^{- 1} {\frac{1}{( s + \frac{5}{2} ) ^{2} + \frac{19}{4}}} : e^{- \frac{5 t}{2}} \frac{2}{19} sin (\frac{19 t}{2})

= \frac{9}{5} e^{- \frac{5 t}{2}} cos (\frac{19 t}{2}) - \frac{9}{2} e^{- \frac{5 t}{2}} \frac{2}{19} sin (\frac{19 t}{2})

整理

\frac{9}{5} e^{- \frac{5 t}{2}} cos (\frac{19 t}{2}) - \frac{9}{2} e^{- \frac{5 t}{2}} \frac{2}{19} sin (\frac{19 t}{2}) : \frac{9}{5} e^{- \frac{5 t}{2}} cos (\frac{19 t}{2}) - \frac{9 e ^{- \frac{5 t}{2}} sin ( \frac{19 t}{2} )}{19}

= \frac{9}{5} e^{- \frac{5 t}{2}} cos (\frac{19 t}{2}) - \frac{9 e ^{- \frac{5 t}{2}} sin ( \frac{19 t}{2} )}{19}

d er i v a t i v e \frac{x}{x - 1}

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

f (x) = 5 - 3 (1 + x^{2})^{- 1}

x \to 1 lim (\frac{2 x ^{3} - 14 x ^{2} + 12 x}{x ^{3} - 10 x ^{2} + 27 x - 18})

x \cdot \frac{d y}{d x} + y = y^{2} x^{2} ln (x)