laplacetransform e^{4t}(t^2+3t+5)

解答

拉普拉斯变换

e^{4 t} (t^{2} + 3 t + 5)

\frac{5 s ^{2} - 37 s + 70}{( s - 4 ) ^{3}}

求解步骤

L {e^{4 t} (t^{2} + 3 t + 5)}

应用变换法则: if

L {f (t)} = F (s)

则

L {e^{a t} f (t)} = F (s - a)

对于

e^{4 t} (t^{2} + 3 t + 5) : f (t) = (t^{2} + 3 t + 5), a = 4

= L {(t^{2} + 3 t + 5)} (s - 4)

L {(t^{2} + 3 t + 5)} : \frac{2}{s ^{3}} + \frac{3}{s ^{2}} + \frac{5}{s}

= \frac{2}{( s - 4 ) ^{3}} + \frac{3}{( s - 4 ) ^{2}} + \frac{5}{( s - 4 )}

= \frac{2}{( s - 4 ) ^{3}} + \frac{3}{( s - 4 ) ^{2}} + \frac{5}{s - 4}

化简

\frac{2}{( s - 4 ) ^{3}} + \frac{3}{( s - 4 ) ^{2}} + \frac{5}{s - 4} : \frac{5 s ^{2} - 37 s + 70}{( s - 4 ) ^{3}}

= \frac{5 s ^{2} - 37 s + 70}{( s - 4 ) ^{3}}

\int (- 9 x^{2} sec (3 x^{3} + 1)) d x

\frac{d}{d x} ((4 x)^{5 x})

\int (3 x)^{2} + 2 d x

x \to 1 lim (1 - 2 - x)

\frac{d}{d x} (arccos (x^{3}))