Question

what is the inverse laplace transform of log s+3/s-3

Answer 1

Answer:

f(s)=loges+2s−5=loge(s+2)−loge(s−5)

We’d obtain:

f′(s)=df(s)ds=1s+2−1s−5

Now we can easily determine the Inverse Laplace Transforms of 1s+2 and 1s−5, which are e−2t and e5t, respectively. Therefore, differentiating f(s) w.r.t.s is the route we take.

Differentiating f(s) w.r.t.s is equivalent to multiplying F(t) by t. As such, we start from the known result for Multiplying F(t) by t:

L{tF(t)}=−df(s)ds=−f′(s)

∴tF(t)=−L−1{f′(s)}=−L−1{1s+2−1s−5}=−(e−2t−e5t)

=e5t−e−2t

∴F(t)=e5t−e−2tt

Answer 2

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