inverselaplace ((s+3))/(s^2+s+3)

解答

的拉普拉斯逆变换

\frac{( s + 3 )}{s ^{2} + s + 3}

e^{- \frac{t}{2}} cos (\frac{11 t}{2}) + \frac{5 e ^{- \frac{t}{2}} sin ( \frac{11 t}{2} )}{11}

求解步骤

L^{- 1} {\frac{( s + 3 )}{s ^{2} + s + 3}}

展开

\frac{s + 3}{s ^{2} + s + 3} : \frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{11}{4}} + \frac{5}{2} \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{11}{4}}

= L^{- 1} {\frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{11}{4}} + \frac{5}{2} \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{11}{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{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{11}{4}}} + \frac{5}{2} L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{11}{4}}}

L^{- 1} {\frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{11}{4}}} : e^{- \frac{t}{2}} cos (\frac{11 t}{2})

L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{11}{4}}} : e^{- \frac{t}{2}} \frac{2}{11} sin (\frac{11 t}{2})

= e^{- \frac{t}{2}} cos (\frac{11 t}{2}) + \frac{5}{2} e^{- \frac{t}{2}} \frac{2}{11} sin (\frac{11 t}{2})

整理

e^{- \frac{t}{2}} cos (\frac{11 t}{2}) + \frac{5}{2} e^{- \frac{t}{2}} \frac{2}{11} sin (\frac{11 t}{2}) : e^{- \frac{t}{2}} cos (\frac{11 t}{2}) + \frac{5 e ^{- \frac{t}{2}} sin ( \frac{11 t}{2} )}{11}

= e^{- \frac{t}{2}} cos (\frac{11 t}{2}) + \frac{5 e ^{- \frac{t}{2}} sin ( \frac{11 t}{2} )}{11}

\frac{d}{d x} \frac{d y}{d x} - a^{2} y + b = 0

x \to \infty lim (\frac{45}{x})

a re a y = x + 1, y = 0, x = 0, x = 2

\frac{d}{d θ} (cos (2 θ))

\frac{d}{d x} (\frac{1}{3} \cdot x^{3})