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
ANSWERED
asked 2021-03-07
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..

Answers (1)

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:
image

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