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Homework Statement
Let
##f(t)=
\begin{cases}
\sin t , \; \; 0 \le t < \pi \\
0 , \; \; \; \; \; \text{else.}
\end{cases}##
Use Laplace transform to solve the initial value problem
##x'(t)+x(t)=f(t), \; \; \; x(0)=0.##
Homework Equations
Some useful Laplace transforms:
##\mathscr{L}\left(f'(t)\right) = z\mathscr{L}\left(f(z)\right)-f(0)##
##\mathscr{L}\left(H(t-a)f(t-a)\right) = e^{-az}\mathscr{L}\left(f(z) \right)##
for ##a >0## and ##H(t)## is the Heaviside function. (i.e. we shift the function)
##\mathscr{L}\left( \sin ct\right) = \frac{c}{z^2+c^2}##
##\mathscr{L}\left(t^ve^{ct}\right) = \frac{\Gamma (v+1)}{(z-c)^{v+1}}##
for ##v > -1## and ##\Gamma## is the gamma function.
The Attempt at a Solution
We write ##f(t) = (H(t)-H(t-\pi))\sin t##. Transforming the ODE we get
##z\mathscr{L}\left(x(t)\right)-x(0)+\mathscr{L}\left(x(t)\right)=\mathscr{L}\left(f(t)\right) ##
Rearranging (We don't actually need compute the Laplace transform here I think as long as we're convinced it does exist.)
##\mathscr{L}\left( x(t) \right) = \frac{1}{z+1}\mathscr{L}\left(f(t)\right)##
Taking the inverse Laplace transform we have
##x(t) = (e^{-t})*((H(t)-H(t-\pi)\sin t)##
Computing the convolution we have
##x(t) = \int_{-\infty}^\infty \left(e^{y-t}(H(-y)-H(-y+\pi)\sin -y \right)dy = \int_{-\infty}^0 -e^{y-t}\sin y dy + \int_{-\infty}^\pi e^{y-t}\sin y dy = \frac{e^{-t}}{2} + \frac{e^{-t}+\pi}{2} = \frac{e^{-t}}{2}(1+e^{\pi})##
Not entirely sure what I'm doing wrong here. The solution should be
##x(t) = \begin{cases}
\frac{e^{-t}-\cos t + \sin t}{2} \; \; t < 0 < \pi\\
\frac{e^{-t}}{2}(1+e^{\pi}) \; \; t > \pi
\end{cases}##
The solution for ##t > \pi## looks suspiciously alike to my solution.
On the other hand if we calculate the Laplace transform earlier we have
##\mathscr{L}\left( H(t-\pi)\sin(t) \right) = \mathscr{L}\left( -H(t-\pi)\sin(t-\pi) \right) = \frac{-e^{\pi z}}{z^2+1}## so we have
##\mathscr{L}\left( f(t) \right) = \frac{1-e^{\pi z}}{(z^2+1)(z+1)}##
Interesting enough the first part ##\frac{1}{(z^2+1)(z+1)}## has the Inverse Laplace transform
##\frac{e^{-t}-\cos t + \sin t}{2}## which also looks similar.
I kept out most of the integral calculations to make the post a bit shorter, they're all pretty much integration by parts twice. I also checked them with wolfram alpha to make sure they're correct.
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