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受欢迎的微积分 >

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

解答

的拉普拉斯逆变换

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

解答

\frac{5}{2} H (t) - \frac{5}{2} e^{- \frac{t}{2}} cos (\frac{7 t}{2}) - \frac{5 e ^{- \frac{t}{2}} sin ( \frac{7 t}{2} )}{2 7}

求解步骤

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

将

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

用部份分式展开:

\frac{5}{2 s} + \frac{- 5 s - 5}{2 ( s ^{2} + s + 2 )}

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

展开

\frac{- 5 s - 5}{2 ( s ^{2} + s + 2 )} : - \frac{5}{2} \cdot \frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{7}{4}} - \frac{5}{4} \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{7}{4}}

= L^{- 1} {\frac{5}{2 s} - \frac{5}{2} \cdot \frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{7}{4}} - \frac{5}{4} \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{7}{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)}

= L^{- 1} {\frac{5}{2 s}} - \frac{5}{2} L^{- 1} {\frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{7}{4}}} - \frac{5}{4} L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{7}{4}}}

L^{- 1} {\frac{5}{2 s}} : \frac{5}{2} H (t)

L^{- 1} {\frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{7}{4}}} : e^{- \frac{t}{2}} cos (\frac{7 t}{2})

L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{7}{4}}} : e^{- \frac{t}{2}} \frac{2}{7} sin (\frac{7 t}{2})

= \frac{5}{2} H (t) - \frac{5}{2} e^{- \frac{t}{2}} cos (\frac{7 t}{2}) - \frac{5}{4} e^{- \frac{t}{2}} \frac{2}{7} sin (\frac{7 t}{2})

整理

\frac{5}{2} H (t) - \frac{5}{2} e^{- \frac{t}{2}} cos (\frac{7 t}{2}) - \frac{5}{4} e^{- \frac{t}{2}} \frac{2}{7} sin (\frac{7 t}{2}) : \frac{5}{2} H (t) - \frac{5}{2} e^{- \frac{t}{2}} cos (\frac{7 t}{2}) - \frac{5 e ^{- \frac{t}{2}} sin ( \frac{7 t}{2} )}{2 7}

= \frac{5}{2} H (t) - \frac{5}{2} e^{- \frac{t}{2}} cos (\frac{7 t}{2}) - \frac{5 e ^{- \frac{t}{2}} sin ( \frac{7 t}{2} )}{2 7}

流行的例子

(dy)/(dx)y=13^x+e^2

\frac{d y}{d x} y = 1 3^{x} + e^{2}

integral from-pi/6 to 0 of 4sec^3(x)

\int_{- \frac{π}{6}}^{0} 4 sec^{3} (x) d x

limit as h approaches 0 of (e^h-1)/h

h \to 0 lim (\frac{e ^{h} - 1}{h})

(\partial)/(\partial w)(ln(2w^3-7y+3xy^2))

\frac{\partial}{\partial w} (ln (2 w^{3} - 7 y + 3 x y^{2}))

奇偶性 f(x)=cot^3(sin(x))

p a r i t y f (x) = cot^{3} (sin (x))