laplacetransform t^2(8e^{2t}-6sin(t))

解答

拉普拉斯变换

t^{2} (8 e^{2 t} - 6 sin (t))

- \frac{4 ( - 4 s ^{6} + 9 s ^{5} - 66 s ^{4} + 105 s ^{3} - 66 s ^{2} - 36 s + 20 )}{( s - 2 ) ^{3} ( s ^{2} + 1 ) ^{3}}

求解步骤

L {t^{2} (8 e^{2 t} - 6 sin (t))}

使用拉普拉斯变换表:

L {t^{k} f (t)} = (- 1)^{k} \frac{d ^{k}}{d s ^{k}} (L {f (t)})

对于

t^{2} (8 e^{2 t} - 6 sin (t)) : f (t) = (8 e^{2 t} - 6 sin (t)), k = 2

= (- 1)^{2} \frac{d ^{2}}{d s ^{2}} (L {(8 e^{2 t} - 6 sin (t))})

L {(8 e^{2 t} - 6 sin (t))} : \frac{8}{s - 2} - \frac{6}{s ^{2} + 1}

\frac{d ^{2}}{d s ^{2}} (\frac{8}{s - 2} - \frac{6}{s ^{2} + 1}) = - \frac{4 ( - 4 s ^{6} + 9 s ^{5} - 66 s ^{4} + 105 s ^{3} - 66 s ^{2} - 36 s + 20 )}{( s - 2 ) ^{3} ( s ^{2} + 1 ) ^{3}}

= (- 1)^{2} (- \frac{4 ( - 4 s ^{6} + 9 s ^{5} - 66 s ^{4} + 105 s ^{3} - 66 s ^{2} - 36 s + 20 )}{( s - 2 ) ^{3} ( s ^{2} + 1 ) ^{3}})

= - \frac{4 ( - 4 s ^{6} + 9 s ^{5} - 66 s ^{4} + 105 s ^{3} - 66 s ^{2} - 36 s + 20 )}{( s - 2 ) ^{3} ( s ^{2} + 1 ) ^{3}}

x \to 0 lim ((1 + 4 x)^{\frac{4}{x}})

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\frac{d}{d x} (10 e^{- 2 x})