laplacetransform cos(t)*e^t

解答

拉普拉斯变换

cos (t) \cdot e^{t}

\frac{s - 1}{( s - 1 ) ^{2} + 1}

求解步骤

L {e^{t} cos (t)}

应用变换法则: if

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

则

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

对于

cos (t) e^{t} : f (t) = cos (t), a = 1

= L {cos (t)} (s - 1)

L {cos (t)} : \frac{s}{s ^{2} + 1}

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

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

\int \frac{tan ^{3} ( x )}{sec ^{2} ( x ) - 1} d x

d er i v a t i v e y = ln (x ln (x))

\int sec^{5} (t) d t

\frac{\partial ^{2}}{\partial x ^{2}} (x^{2} - e^{y^{2}})

d er i v a t i v e ln ∣ sec (θ) + tan (θ) ∣