laplacetransform 3e^t(32t^2+2t+15)

解答

拉普拉斯变换

3 e^{t} (32 t^{2} + 2 t + 15)

\frac{3 ( 15 s ^{2} - 28 s + 77 )}{( s - 1 ) ^{3}}

求解步骤

L {3 e^{t} (32 t^{2} + 2 t + 15)}

利用拉普拉斯变换的常数乘法特性:

对于函数

f (t)

和常数

a : L {a \cdot f (t)} = a \cdot L {f (t)}

= 3 L {e^{t} (32 t^{2} + 2 t + 15)}

L {e^{t} (32 t^{2} + 2 t + 15)} : \frac{64}{( s - 1 ) ^{3}} + \frac{2}{( s - 1 ) ^{2}} + \frac{15}{s - 1}

= 3 (\frac{64}{( s - 1 ) ^{3}} + \frac{2}{( s - 1 ) ^{2}} + \frac{15}{s - 1})

化简

3 (\frac{64}{( s - 1 ) ^{3}} + \frac{2}{( s - 1 ) ^{2}} + \frac{15}{s - 1}) : \frac{3 ( 15 s ^{2} - 28 s + 77 )}{( s - 1 ) ^{3}}

= \frac{3 ( 15 s ^{2} - 28 s + 77 )}{( s - 1 ) ^{3}}

\int \frac{3}{e ^{q} + 2} d q

\int e^{\frac{R}{L} t} d t

d er i v a t i v e f (x) = 4 x - 3 x + 7

\int \frac{1}{x ^{2} + 2 x + 101} d x

\int \frac{2}{2 - x ^{2}} d x