# 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..

Question
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 $${u}_{\pi}{\left({t}\right)}={\left\lbrace\begin{matrix}{1}&{t}&\ge&\pi\\{0}&{t}&<&\pi\end{matrix}\right.},$$</span>
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 $${t}<\pi,{t}{h}{e}{n}{y}{\left({t}\right)}=\frac{1}{{3}}{\left[ \sin{{t}}-\frac{1}{{2}} \sin{{\left({2}{t}\right)}}\right]}{\quad\text{and}\quad} f{{\left({t}\right)}}= \sin{{t}}.$$</span>
The graph of the solution and of the forcing function is:

### Relevant Questions

To determine the solution of the initial value problem
$${y}{''}+{4}{y}=\delta{\left({t}-\pi\right)}-\delta{\left({t}-{2}\pi\right)},{y}{\left({0}\right)}={0},{y}'{\left({0}\right)}={0}$$
and draw the graph of the solution.
The solution of the initial value problem $$y''\ +\ 4y=2\delta(t\ -\ \pi/4),\ y(0)=0,\ y'(0)=0$$ and draw the graph of the solution.
The solution of initial value problem $$y''\ +\ 2y'\ +\ 2y=\delta(t\ -\ \pi),\ y(0)=0,\ y'(0)=1$$ and draw the graph of the solution.
To state:
The solution of the given initial value problem using the method of Laplace transforms.
Given:
The initial value problem is,
$${y}\text{}+{6}{y}'+{5}{y}={12}{e}^{t},{y}{\left({0}\right)}=-{1},{y}'{\left({0}\right)}={7}$$
To state:
The solution of the given initial value problem using the method of Laplace transforms.
Given:
The initial value problem is,
$${y}\text{}+{2}{y}'+{5}{y}={0},{y}{\left({0}\right)}={2},{y}'{\left({0}\right)}={4}$$
a) To graph: the $${k}{e}{r}{\left({A}\right)},{\left({k}{e}{r}{A}\right)}^{\bot}{\quad\text{and}\quad}{i}{m}{\left({A}^{T}\right)}$$
b) To find: the relationship between im $$(A^{T})$$ and ker (A).
c) To find: the relationship between ker(A) and solution set S
d) To find vecx_0 at the intersection of $${k}{e}{r}{\left({A}\right)}{\quad\text{and}\quad}{\left({k}{e}{r}{A}\right)}^{\bot}$$
e) To find: the lengths of $$\vec{{x}}_{{0}}$$ compared to the other vectors in S
In an integro-differential equation, the unknown dependent variable y appears within an integral, and its derivative $$\frac{dy}{dt}$$ also appears. Consider the following initial value problem, defined for t > 0:
$$\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}+{4}{\int_{{0}}^{{t}}}{y}{\left({t}-{w}\right)}{e}^{{-{4}{w}}}{d}{w}={3},{y}{\left({0}\right)}={0}$$
a) Use convolution and Laplace transforms to find the Laplace transform of the solution.
$${Y}{\left({s}\right)}={L}{\left\lbrace{y}{\left({t}\right)}\right)}{\rbrace}-?$$
b) Obtain the solution y(t).
y(t) - ?
$$4''\ +\ 4y'4y=\delta(t\ -\ 4)$$
$$y(0) = 1$$
$$y'(0) = 2$$
$${T}:{M}_{{3.3}},$$
$${T}{\left({A}\right)}={\left[\begin{matrix}{0}&{0}&{1}\\{0}&{1}&{0}\\{1}&{0}&{0}\end{matrix}\right]}{A}$$
Solve the initial-value problem on the specified interval $${y}'+{y} \tan{{x}}= \sin{{2}}{x}{o}{n}{\left(-\frac{1}{{2}}\pi,\frac{1}{{2}}\pi\right)}\ \text{with}\ {y}={2}\ \text{when}\ {x}={0}$$.