Question

# 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..

Transformation properties
To determine the solution of the initial value problem
$${y}{''}+{4}{y}= \sin{{t}}+{u}_{\pi}{\left({t}\right)} \sin{{\left({t}-\pi\right)}}:$$
$$y(0) = 0,$$
$$y'(0) = 0.$$
Also, draw the graphs of the solution and of the forcing function and explain the relation between the solution and the forcing function..

2021-03-08

Applying the Laplace transform to both sides of the differential equation
$${y}{''}+{4}{y}= \sin{{t}}+{u}_{\pi}{\left({t}\right)} \sin{{\left({t}-\pi\right)}}.$$
$$\Rightarrow{L}{\left\lbrace{y}{''}+{4}{y}\right\rbrace}={L}{\left\lbrace \sin{{t}}+{u}_{\pi}{\left({t}\right)} \sin{{\left({t}-\pi\right)}}\right\rbrace}$$
$$\Rightarrow{L}{\left\lbrace{y}{''}\right\rbrace}+{L}{\left\lbrace{4}{y}\right\rbrace}={L}{\left\lbrace \sin{{t}}\right\rbrace}+{L}{\left\lbrace{u}_{\pi}{\left({t}{0} \sin{{\left({t}-\pi\right)}}\right\rbrace}\right.}$$
By using $${L}{\left\lbrace{{f}^{n}{\left({t}\right)}}\right\rbrace}={s}^{n}{F}{\left({s}\right)}-{s}^{{{n}-{1}}}{\left({0}\right)}-\ldots-{{f}^{{{\left({n}-{1}\right)}}}{\left({0}\right)}}$$ and
$${e}^{{-{c}{s}}}{G}{\left({s}\right)}={L}{\left\lbrace{u}_{{c}}{\left({t}\right)} g{{\left({t}-{c}\right)}}\right\rbrace},{L}{\left\lbrace \sin{{t}}\right\rbrace}=\frac{1}{{{s}^{2}+{1}}}$$
$$\Rightarrow{\left[{s}^{2}{Y}{\left({s}\right)}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}\right]}+{4}{Y}{\left({s}\right)}=\frac{1}{{{s}^{2}+{1}}}+\frac{{e}^{{-\pi{s}}}}{{{s}^{2}+{1}}}$$
Using the initial conditions, $$y(0) = 0\ \text{and}\ y'(0) = 0,$$
$$\Rightarrow{s}^{2}{Y}{\left({s}\right)}+{4}{Y}{\left({s}\right)}=\frac{{{1}+{e}^{{-\pi{s}}}}}{{{s}^{2}+{1}}}$$
$$\Rightarrow{Y}{\left({s}\right)}{\left[{s}^{2}+{4}\right]}=\frac{{{1}+{e}^{{-\pi{s}}}}}{{{s}^{2}+{1}}}$$
$$\Rightarrow{Y}{\left({s}\right)}=\frac{{{1}+{e}^{{-\pi{s}}}}}{{{\left({s}^{2}+{1}\right)}{\left({s}^{2}+{4}\right)}}}$$
By using the partial fraction $$\frac{{{1}+{e}^{{-\pi{s}}}}}{{{\left({s}^{2}+{1}\right)}{\left({s}^{2}+{4}\right)}}}=\frac{1}{{3}}{\left({1}+{e}^{{-\pi{s}}}\right)}\lfloor\frac{1}{{{\left({s}^{2}+{1}\right)}}}+\frac{1}{{{\left({s}^{2}+{4}\right)}}}\rfloor$$
$$\Rightarrow{Y}{\left({s}\right)}=\frac{1}{{3}}{\left({1}+{e}^{{-\pi{s}}}\right)}\lfloor\frac{1}{{{\left({s}^{2}+{1}\right)}}}+\frac{1}{{{\left({s}^{2}+{4}\right)}}}\rfloor$$
Applying the inverse Laplace transformation on both sides,
$$\Rightarrow{L}^{ -{{1}}}{\left\lbrace{Y}{\left({s}\right)}\right\rbrace}={L}^{ -{{1}}}{\left\lbrace\frac{1}{{3}}{\left({1}+{e}^{{-\pi{s}}}\right)}\lfloor\frac{1}{{{\left({s}^{2}+{1}\right)}}}+\frac{1}{{{\left({s}^{2}+{4}\right)}}}\rfloor\right\rbrace}$$
$$\Rightarrow{L}^{ -{{1}}}{\left\lbrace{Y}{\left({s}\right)}\right\rbrace}={L}^{ -{{1}}}{\left\lbrace\frac{1}{{3}}\lfloor\frac{{{\left({1}+{e}^{{-\pi{s}}}\right)}}}{{{\left({s}^{2}+{1}\right)}}}+\frac{{{\left({1}+{e}^{{-\pi{s}}}\right)}}}{{{\left({s}^{2}+{4}\right)}}}\rfloor\right\rbrace}$$
$$\Rightarrow{L}^{ -{{1}}}{\left\lbrace{Y}{\left({s}\right)}\right\rbrace}=\frac{1}{{3}}{L}^{ -{{1}}}{\left\lbrace{\left[\frac{1}{{{s}^{2}+{1}}}+\frac{{e}^{{-\pi{s}}}}{{{s}^{2}+{1}}}-\frac{1}{{{s}^{2}+{4}}}-\frac{{e}^{{-\pi{s}}}}{{{s}^{2}+{4}}}\right]}\right\rbrace}$$
$$\Rightarrow{L}^{ -{{1}}}{\left\lbrace{Y}{\left({s}\right)}\right\rbrace}=\frac{1}{{3}}{\left[{L}^{ -{{1}}}{\left\lbrace\frac{1}{{{s}^{2}+{1}}}\right\rbrace}+{L}^{ -{{1}}}{\left\lbrace\frac{{e}^{{-\pi{s}}}}{{{s}^{2}+{1}}}\right\rbrace}-{L}^{ -{{1}}}{\left\lbrace\frac{1}{{{s}^{2}+{4}}}\right\rbrace}-{L}^{ -{{1}}}{\left\lbrace\frac{{e}^{{-\pi{s}}}}{{{s}^{2}+{4}}}\right\rbrace}\right]}$$
By using $${L}^{ -{{1}}}{\left\lbrace{e}^{{-{c}{s}}}{G}{\left({s}\right)}\right\rbrace}={u}_{{s}}{\left({t}\right)} g{{\left({t}-{c}\right)}}$$ and the properties of inverse Laplace,
$$\Rightarrow{y}{\left({t}\right)}=\frac{1}{{3}}{\left[ \sin{{t}}+{u}_{\pi}{\left({t}\right)} \sin{{\left({t}-\pi\right)}}-\frac{1}{{2}} \sin{{\left({2}{t}\right)}}-\frac{1}{{2}}{u}_{\pi}{\left({t}\right)} \sin{{\left({2}{\left({t}-\pi\right)}\right)}}\right]}$$
Thus $${y}{\left({t}\right)}=\frac{1}{{3}}{\left[ \sin{{t}}+{u}_{\pi}{\left({t}\right)} \sin{{\left({t}-\pi\right)}}-\frac{1}{{2}} \sin{{\left({2}{t}\right)}}-\frac{1}{{2}}{u}_{\pi}{\left({t}\right)} \sin{{\left({2}{\left({t}-\pi\right)}\right)}}\right]}$$
Therefore, the solution of the initial value problem
$${y}{''}+{4}{y}= \sin{{t}}+{u}_{\pi}{\left({t}\right)} \sin{{\left({t}-\pi\right)}},$$
$$y(0) = 0,$$
$$y'(0) = 0$$ is
$${y}{\left({t}\right)}=\frac{1}{{3}}{\left[ \sin{{t}}+{u}_{\pi}{\left({t}\right)} \sin{{\left({t}-\pi\right)}}-\frac{1}{{2}} \sin{{\left({2}{t}\right)}}-\frac{1}{{2}}{u}_{\pi}{\left({t}\right)} \sin{{2}}{\left({\left({t}-\pi\right)}\right)}\right]}$$
Here, the forcing function is $$f{{\left({t}\right)}}= \sin{{t}}+{u}_{\pi}{\left({t}\right)} \sin{{\left({t}-\pi\right)}}$$
By using $${u}_{\pi}{\left({t}\right)}={\left\lbrace\begin{matrix}{1}&{t}&\ge&\pi\\{0}&{t}&<&\pi\end{matrix}\right.},$$
For $${t}\ge\pi\ {t}{h}{e}{n}{y}\ {\left({t}\right)}=\frac{1}{{3}}{\left[ \sin{{t}}+ \sin{{\left({t}-\pi\right)}}-\frac{1}{{2}} \sin{{\left({2}{t}\right)}}-\frac{1}{{2}} \sin{{\left({2}{\left({t}-\pi\right)}\right)}}\right]}$$ and
$$f{{\left({t}\right)}}= \sin{{t}}+ \sin{{\left({t}-\pi\right)}}.$$
For $${t}<\pi,\ {t}{h}{e}{n}\ {y}{\left({t}\right)}=\frac{1}{{3}}{\left[ \sin{{t}}-\frac{1}{{2}} \sin{{\left({2}{t}\right)}}\right]}{\quad\text{and}\quad} f{{\left({t}\right)}}= \sin{{t}}.$$
The graph of the solution and of the forcing function is: