laplacetransform 2(t-2)^2+6(t-2)+6

解答

拉普拉斯变换

2 (t - 2)^{2} + 6 (t - 2) + 6

\frac{4}{s ^{3}} - \frac{2}{s ^{2}} + \frac{2}{s}

求解步骤

L {2 (t - 2)^{2} + 6 (t - 2) + 6}

展开

2 (t - 2)^{2} + 6 (t - 2) + 6 : 2 t^{2} - 2 t + 2

= L {2 t^{2} - 2 t + 2}

利用拉普拉斯变换的线性特性:

对于函数

f (t), g (t)

和常数

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= 2 L {t^{2}} - 2 L {t} + L {2}

L {t^{2}} : \frac{2}{s ^{3}}

L {t} : \frac{1}{s ^{2}}

L {2} : \frac{2}{s}

= 2 \cdot \frac{2}{s ^{3}} - 2 \cdot \frac{1}{s ^{2}} + \frac{2}{s}

整理

2 \frac{2}{s ^{3}} - 2 \frac{1}{s ^{2}} + \frac{2}{s} : \frac{4}{s ^{3}} - \frac{2}{s ^{2}} + \frac{2}{s}

= \frac{4}{s ^{3}} - \frac{2}{s ^{2}} + \frac{2}{s}

\int \frac{1}{4 x ^{2} - 16} d x

\int \frac{2}{x ^{2} - 4 x + 13} d x

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