inverselaplace s/(10(s^2+10s+9.8))

解答

的拉普拉斯逆变换

\frac{s}{10 ( s ^{2} + 10 s + 9.8 )}

0.1 e^{- 5 t} cosh (3.89871 \dots t) - 0.12824 \dots e^{- 5 t} sinh (3.89871 \dots t)

求解步骤

L^{- 1} {\frac{s}{10 ( s ^{2} + 10 s + 9.8 )}}

展开

\frac{s}{10 ( s ^{2} + 10 s + 9.8 )} : \frac{1}{10} \cdot \frac{s + 5}{( s + 5 ) ^{2} - 15.2} - \frac{1}{2} \cdot \frac{1}{( s + 5 ) ^{2} - 15.2}

= L^{- 1} {\frac{1}{10} \cdot \frac{s + 5}{( s + 5 ) ^{2} - 15.2} - \frac{1}{2} \cdot \frac{1}{( s + 5 ) ^{2} - 15.2}}

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{1}{10} L^{- 1} {\frac{s + 5}{( s + 5 ) ^{2} - 15.2}} - \frac{1}{2} L^{- 1} {\frac{1}{( s + 5 ) ^{2} - 15.2}}

L^{- 1} {\frac{s + 5}{( s + 5 ) ^{2} - 15.2}} : e^{- 5 t} cosh (3.89871 \dots t)

L^{- 1} {\frac{1}{( s + 5 ) ^{2} - 15.2}} : e^{- 5 t} \frac{1}{3.89871 \dots} sinh (3.89871 \dots t)

= \frac{1}{10} e^{- 5 t} cosh (3.89871 \dots t) - \frac{1}{2} e^{- 5 t} \frac{1}{3.89871 \dots} sinh (3.89871 \dots t)

化简

\frac{1}{10} e^{- 5 t} cosh (3.89871 \dots t) - \frac{1}{2} e^{- 5 t} \frac{1}{3.89871 \dots} sinh (3.89871 \dots t) : 0.1 e^{- 5 t} cosh (3.89871 \dots t) - 0.12824 \dots e^{- 5 t} sinh (3.89871 \dots t)

= 0.1 e^{- 5 t} cosh (3.89871 \dots t) - 0.12824 \dots e^{- 5 t} sinh (3.89871 \dots t)

\frac{d}{d x} ((x)^{3})

\int x^{7} x^{4} + 7 d x

d er i v a t i v e \frac{4 x + 4}{3 x ^{\frac{2}{3}}}

\frac{d}{d x} (cos^{6} (3 x^{2} + 4))

l a pl a ce t r an s f or m 3 sin (2)