Find
𝐿
{
𝑡
𝑒
𝑡
cos
⁡
(
6
𝑡
)
}

Question

Find 
𝐿
{
𝑡
𝑒
𝑡
cos
⁡
(
6
𝑡
)
}

BlackTom AI · Accepted Answer

The problem asks for the Laplace transform of t·e^{t}·cos(6t). A good way to tackle this is to use a standard Laplace transform property: L{t·f(t)} = -d/ds [L{f(t)}]. First, identify f(t) = e^{t}·cos(6t). Recall the shift-in-time in the exponential: L{e^{a t}·cos(b t)} = (s - a) / [(s - a)^2 + b^2]. Here, a = 1 and b = 6, so L{e^{t}·cos(6t)} = (s - 1) / [(s - 1)^2 + 36].

Now apply the t-multiplication property: L{t·e^{t}·cos(6t)} = -d/ds { (s - 1) / [(s - 1)^2 + 36] }.

Let u = s - 1 for simplicity. Then F(s) = u / (u^2 + 36). Differentiate with respect to s (note du/ds = 1): dF/ds = d/du [ u / (u^2 + 36) ] = [ (u^2 + 36)·1 − u·(2u) ] / (u^2 + 36)^2 = (36 − u^2) / (u^2 + 36)^2.

Therefore, L{t·e^{t}·cos(6t)} = −dF/ds = −(36 − u^2)/(u^2 + 36)^2 = (u^2 − 36)/(u^2 + 36)^2.

Substitute back u = s − 1: the final result is ((s − 1)^2 − 36) / [ (s − 1)^2 + 36 ]^2.

The given option matches this expression, since it is precisely ((s − 1)^2 − 36) divided by ((s − 1)^2 + 36) squared.

In summary, the correct form is ((s − 1)^2 − 36) / ((s − 1)^2 + 36)^2, which results from applying the Laplace transform of e^{t}cos(6t) and the t-multiplication property.

Find 𝐿 { 𝑡 𝑒 𝑡 cos ⁡ ( 6 𝑡 ) } 单项选择题

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