inverselaplace (10)/(16s^2+5s+1)

解答

的拉普拉斯逆变换

\frac{10}{16 s ^{2} + 5 s + 1}

\frac{20 e ^{- \frac{5 t}{32}} sin ( \frac{39 t}{32} )}{39}

求解步骤

L^{- 1} {\frac{10}{16 s ^{2} + 5 s + 1}}

展开

\frac{10}{16 s ^{2} + 5 s + 1} : \frac{5}{8} \cdot \frac{1}{( s + \frac{5}{32} ) ^{2} + \frac{39}{1024}}

= L^{- 1} {\frac{5}{8} \cdot \frac{1}{( s + \frac{5}{32} ) ^{2} + \frac{39}{1024}}}

利用逆拉普拉斯变换的常数乘法特性:

对于函数

f (t)

和常数

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= \frac{5}{8} L^{- 1} {\frac{1}{( s + \frac{5}{32} ) ^{2} + \frac{39}{1024}}}

L^{- 1} {\frac{1}{( s + \frac{5}{32} ) ^{2} + \frac{39}{1024}}} : e^{- \frac{5 t}{32}} \frac{32}{39} sin (\frac{39 t}{32})

= \frac{5}{8} e^{- \frac{5 t}{32}} \frac{32}{39} sin (\frac{39 t}{32})

整理

\frac{5}{8} e^{- \frac{5 t}{32}} \frac{32}{39} sin (\frac{39 t}{32}) : \frac{20 e ^{- \frac{5 t}{32}} sin ( \frac{39 t}{32} )}{39}

= \frac{20 e ^{- \frac{5 t}{32}} sin ( \frac{39 t}{32} )}{39}

\frac{\partial}{\partial x} (x^{2} + 4 y^{2} + 5 z^{2})

f (x) = \frac{7}{10} x^{2} + 4 x^{3}

\frac{d p}{p - p ^{2}} = d t

\frac{d}{d x} (3 x^{4} + 2 x^{3} + x^{2} - 2)

\frac{d ^{2}}{d x ^{2}} (x^{- 2})