 # To determine the solution of the initial value problem {y}{''}+{4}{y}= sin{{t}}+{u}_{pi}{left({t}right)} sin{{left({t}-piright)}}: y(0) = 0, y'(0) = 0. Also, draw the graphs of the solution and of the forcing function and explain the relation between the solution and the forcing function.. Anish Buchanan 2021-03-07 Answered
To determine the solution of the initial value problem
$y{}^{″}+4y=\mathrm{sin}t+{u}_{\pi }\left(t\right)\mathrm{sin}\left(t-\pi \right):$
$y\left(0\right)=0,$
${y}^{\prime }\left(0\right)=0.$
Also, draw the graphs of the solution and of the forcing function and explain the relation between the solution and the forcing function..
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Applying the Laplace transform to both sides of the differential equation
$y{}^{″}+4y=\mathrm{sin}t+{u}_{\pi }\left(t\right)\mathrm{sin}\left(t-\pi \right).$
$⇒L\left\{y{}^{″}+4y\right\}=L\left\{\mathrm{sin}t+{u}_{\pi }\left(t\right)\mathrm{sin}\left(t-\pi \right)\right\}$
$⇒L\left\{y{}^{″}\right\}+L\left\{4y\right\}=L\left\{\mathrm{sin}t\right\}+L\left\{{u}_{\pi }\left(t0\mathrm{sin}\left(t-\pi \right)\right\}$
By using $L\left\{{f}^{n}\left(t\right)\right\}={s}^{n}F\left(s\right)-{s}^{n-1}\left(0\right)-\dots -{f}^{\left(n-1\right)}\left(0\right)$ and
${e}^{-cs}G\left(s\right)=L\left\{{u}_{c}\left(t\right)g\left(t-c\right)\right\},L\left\{\mathrm{sin}t\right\}=\frac{1}{{s}^{2}+1}$
$⇒\left[{s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)\right]+4Y\left(s\right)=\frac{1}{{s}^{2}+1}+\frac{{e}^{-\pi s}}{{s}^{2}+1}$
Using the initial conditions,
$⇒{s}^{2}Y\left(s\right)+4Y\left(s\right)=\frac{1+{e}^{-\pi s}}{{s}^{2}+1}$
$⇒Y\left(s\right)\left[{s}^{2}+4\right]=\frac{1+{e}^{-\pi s}}{{s}^{2}+1}$
$⇒Y\left(s\right)=\frac{1+{e}^{-\pi s}}{\left({s}^{2}+1\right)\left({s}^{2}+4\right)}$
By using the partial fraction $\frac{1+{e}^{-\pi s}}{\left({s}^{2}+1\right)\left({s}^{2}+4\right)}=\frac{1}{3}\left(1+{e}^{-\pi s}\right)⌊\frac{1}{\left({s}^{2}+1\right)}+\frac{1}{\left({s}^{2}+4\right)}⌋$
$⇒Y\left(s\right)=\frac{1}{3}\left(1+{e}^{-\pi s}\right)⌊\frac{1}{\left({s}^{2}+1\right)}+\frac{1}{\left({s}^{2}+4\right)}⌋$
Applying the inverse Laplace transformation on both sides,