Question

Laplace Transform Q: http://gyazo.com/51008d0bd2bf3db045c9a9438e4d7d14.png?1338568203 thanks in advance

Answer 1

Notice that the second size can be written
as
\[
\frac{1}{7}
(t-7) (\theta (t-7)-\theta
(t-14))+\theta (t-14)\\ \text { where } (\theta(t-k) \text { is the Heaviside function that is zero before k and one after k}
\]

Answer 2

The Laplace Transform of the second side is
\[
\frac{e^{-7 s} (7 s+1)}{7
s^2}-\frac{e^{-14 s} (14
s+1)}{7 s^2}+\frac{2 e^{-14
s}}{s}-\frac{e^{-7 s}}{s}= \frac{e^{-14 s} \left(e^{7
s}-1\right)}{7 s^2}
\]

Answer 3

The Laplace Transform of the first side is
\[
\left(s^2+9\right) F(s)
\]
Hence
\[
\left(s^2+9\right) F(s)=\frac{e^{-14 s} \left(e^{7
s}-1\right)}{7 s^2}\\
F(s) = \frac{e^{-14 s} \left(e^{7
s}-1\right)}{7 s^2
\left(s^2+9\right)}
\]
Now we have to find the inverse Laplace Transform.

Answer 4

After use of Partial fractions
\[
F(s)=-\frac{e^{-14 s}}{63
s^2}+\frac{e^{-7 s}}{63
s^2}+\frac{e^{-14 s}}{63
\left(s^2+9\right)}-\frac{e
^{-7 s}}{63
\left(s^2+9\right)}
\]

Answer 5

\[
y(t)=L^{-1}(F(s))=\frac{1}{189} (\theta (t-7) (3 (t-7)+\sin
(21-3 t))+\\ \theta (t-14) (-3 t+\sin (3
(t-14))+42))
\]

Answer 6

@experimentX, did you see the solution?

Answer 7

ty prof.!!!