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If

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

and

n = 1, 2, 3, . . . ,

thenscript L{tnf(t)} = (−1)n

dn
dsn
F(s).

Evaluate the given Laplace transform. (Write your answer as a function of s.)

script L{t cos 5t}

User TutuGeorge
by
5.7k points

1 Answer

7 votes

Answer:


(s^2-25)/((s^2+25)^2)

Explanation:

Let's use the definition of the Laplace transform and the identity given:
\mathcal{L}[t \cos 5t]=(-1)F'(s) with
F(s)=\mathcal{L}[\cos 5t].

Now,
F(s)=\int_0 ^(+ \infty)e^(-st)\cos(5t) dt. Using integration by parts with u=e^(-st) and dv=cos(5t), we obtain that
F(s)=(1)/(5)\sin(5t)e^(-st) |_(0)^(+\infty)+(s)/(5)\int_0 ^(+ \infty)e^(-st)\sin(5t) dt=\int_0 ^(+ \infty)e^(-st)\sin(5t) dt.

Using integration by parts again with u=e^(-st) and dv=sin(5t), we obtain that


F(s)=(s)/(5)((-1)/(5)\cos(5t)e^(-st) |_(0)^(+\infty)-(s)/(5)\int_0 ^(+ \infty)e^(-st)\sin(5t) dt)=(s)/(5)((1)/(5)-(s)/(5)\int_0^(+ \infty)e^(-st)\sin(5t) dt)=(s)/(5)-(s^2)/(25)F(s).

Solving for F(s) on the last equation,
F(s)=(s)/(s^2+25), then the Laplace transform we were searching is
-F'(s)=(s^2-25)/((s^2+25)^2)

User DreadedFrost
by
5.6k points