inverselaplace (4s+1)/(2s^2-5s+6)

解答

的拉普拉斯逆变换

\frac{4 s + 1}{2 s ^{2} - 5 s + 6}

2 e^{\frac{5 t}{4}} cos (\frac{23 t}{4}) + \frac{12}{23} e^{\frac{5 t}{4}} sin (\frac{23 t}{4})

求解步骤

L^{- 1} {\frac{4 s + 1}{2 s ^{2} - 5 s + 6}}

展开

\frac{4 s + 1}{2 s ^{2} - 5 s + 6} : 2 \cdot \frac{s - \frac{5}{4}}{( s - \frac{5}{4} ) ^{2} + \frac{23}{16}} + 3 \cdot \frac{1}{( s - \frac{5}{4} ) ^{2} + \frac{23}{16}}

= L^{- 1} {2 \cdot \frac{s - \frac{5}{4}}{( s - \frac{5}{4} ) ^{2} + \frac{23}{16}} + 3 \cdot \frac{1}{( s - \frac{5}{4} ) ^{2} + \frac{23}{16}}}

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)}

= 2 L^{- 1} {\frac{s - \frac{5}{4}}{( s - \frac{5}{4} ) ^{2} + \frac{23}{16}}} + 3 L^{- 1} {\frac{1}{( s - \frac{5}{4} ) ^{2} + \frac{23}{16}}}

L^{- 1} {\frac{s - \frac{5}{4}}{( s - \frac{5}{4} ) ^{2} + \frac{23}{16}}} : e^{\frac{5 t}{4}} cos (\frac{23 t}{4})

L^{- 1} {\frac{1}{( s - \frac{5}{4} ) ^{2} + \frac{23}{16}}} : e^{\frac{5 t}{4}} \frac{4}{23} sin (\frac{23 t}{4})

= 2 e^{\frac{5 t}{4}} cos (\frac{23 t}{4}) + 3 e^{\frac{5 t}{4}} \frac{4}{23} sin (\frac{23 t}{4})

整理

2 e^{\frac{5 t}{4}} cos (\frac{23 t}{4}) + 3 e^{\frac{5 t}{4}} \frac{4}{23} sin (\frac{23 t}{4}) : 2 e^{\frac{5 t}{4}} cos (\frac{23 t}{4}) + \frac{12}{23} e^{\frac{5 t}{4}} sin (\frac{23 t}{4})

= 2 e^{\frac{5 t}{4}} cos (\frac{23 t}{4}) + \frac{12}{23} e^{\frac{5 t}{4}} sin (\frac{23 t}{4})

d er i v a t i v e f (x) = π^{x^{2}}

y^{^{'}} = 2

\int 8 x^{- 5} d x

\int \frac{1}{x ^{2} + 100} d x

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