# use the Laplace transform to solve the given initial-value problem. y"+y=f(t) y(0)=0 , y'(0)=1 where displaystyle f{{left({t}right)}}={leftlbracebegin{matrix}{1}&{0}le{t}<frac{pi}{{2}}{0}&{t}gefrac{pi}{{2}}end{matrix}right.}

Question
Laplace transform
use the Laplace transform to solve the given initial-value problem.
$$y"+y=f(t)$$
$$y(0)=0 , y'(0)=1$$ where
$$\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{1}&{0}\le{t}<\frac{\pi}{{2}}\\{0}&{t}\ge\frac{\pi}{{2}}\end{matrix}\right.}$$</span>

2021-03-05
Step 1
Given, the initial value problem
$$y"+y=f(t)$$
$$y(0)=0 , y'(0)=1$$ where
$$\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{1}&{0}\le{t}<\frac{\pi}{{2}}\\{0}&{t}\ge\frac{\pi}{{2}}\end{matrix}\right.}$$</span>
step 2
The Laplace transform of the two derivatives
$$\displaystyle{L}{\left\lbrace{y}'\right\rbrace}={s}{Y}{\left({s}\right)}-{y}{\left({0}\right)}$$
$$\displaystyle{L}{\left\lbrace{y}\text{}\right\rbrace}={s}^{2}{Y}{\left({s}\right)}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}$$
$$\displaystyle{L}{\left\lbrace{1}\right\rbrace}=\frac{1}{{s}}$$
$$\displaystyle{L}{\left\lbrace{e}^{{{a}{t}}}\right\rbrace}=\frac{1}{{{s}-{a}}}$$
Where y(0) and y’(0) are initial conditions.
Step 3
Now,
when $$\displaystyle{0}\le{t}\le\frac{\pi}{{2}}$$
$$y"+y'=1 , y(0)=0 . y'(0)=1$$
$$\displaystyle{s}^{2}{Y}{\left({s}\right)}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}+{s}{Y}{\left({s}\right)}-{y}{\left({0}\right)}={1}$$
$$\displaystyle{\left\lbrace{s}^{2}+{s}\right\rbrace}{Y}{\left({s}\right)}-{s}\cdot{0}-{1}-{0}={1}$$
$$\displaystyle{\left\lbrace{s}^{2}+{s}\right\rbrace}{Y}{\left({s}\right)}={2}$$
$$\displaystyle{Y}{\left({s}\right)}=\frac{2}{{{s}^{2}+{s}}}=\frac{2}{{{s}{\left({s}+{1}\right)}}}={2}{\left\lbrace\frac{1}{{s}}-\frac{1}{{{s}+{1}}}\right\rbrace}$$
$$\displaystyle{Y}{\left({s}\right)}={2}{\left\lbrace{1}-{e}^{{-{t}}}\right\rbrace}$$
when $$\displaystyle{t}\ge\frac{\pi}{{2}}$$
$$y"+y'=0, y(0)=0 , y'(0)=1$$
$$\displaystyle{s}^{2}{Y}{\left({s}\right)}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}+{s}{Y}{\left({s}\right)}-{y}{\left({0}\right)}={1}$$
$$\displaystyle{\left\lbrace{s}^{2}+{s}\right\rbrace}{Y}{\left({s}\right)}-{s}\cdot{0}-{1}-{0}={0}$$
$$\displaystyle{\left\lbrace{s}^{2}+{s}\right\rbrace}{Y}{\left({s}\right)}={1}$$
$$\displaystyle{Y}{\left({s}\right)}=\frac{1}{{{s}^{2}+{s}}}=\frac{1}{{{s}{\left({s}+{1}\right)}}}={\left\lbrace\frac{1}{{s}}-\frac{1}{{{s}+{1}}}\right\rbrace}$$
$$\displaystyle{Y}{\left({s}\right)}={\left\lbrace{1}-{e}^{{-{t}}}\right\rbrace}$$
Step 4
Hence the solution is
$$\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{2}{\left\lbrace{1}-{e}^{{-{t}}}\right\rbrace}&{0}\le{t}<\frac{\pi}{{2}}\\{1}-{e}^{{-{t}}}&{t}\ge\frac{\pi}{{2}}\end{matrix}\right.}$$</span>

### Relevant Questions

Solve for Y(s), the Laplace transform of the solution y(t) to the initial value problem below.
$$y"+y=g(t) , y(0)=-4 , y'(0)=0$$
where $$g{{\left({t}\right)}}={\left\lbrace\begin{matrix}{t}&{t}<{4}\\{5}&{t}>{4}\end{matrix}\right.}$$
Y(s)-?
Solve the initial value problem $$\displaystyle{\left\lbrace\begin{matrix}{y}\text{}+{16}{y}= \cos{{\left({4}{t}\right)}}\\{y}{\left({0}\right)}={1}\\{y}'{\left({0}\right)}={1}\end{matrix}\right.}$$ using the Laplace transform.
Let f(t) be a function on $$\displaystyle{\left[{0},\infty\right)}$$. The Laplace transform of fis the function F defined by the integral $$\displaystyle{F}{\left({s}\right)}={\int_{{0}}^{\infty}}{e}^{{-{s}{t}}} f{{\left({t}\right)}}{\left.{d}{t}\right.}$$ . Use this definition to determine the Laplace transform of the following function.
$$\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{1}-{t}&{0}<{t}<{1}\\{0}&{1}<{t}\end{matrix}\right.}$$
Use the Laplace transform to solve the given initial-value problem.
$$\displaystyle{y}{''}+{8}{y}'+{41}{y}=\delta{\left({t}-\pi\right)}+\delta{\left({t}-{3}\pi\right)},\ \ \ \ {y}{\left({0}\right)}={1},\ \ \ \ {y}'{\left({0}\right)}={0}\ \ {y}{\left({t}\right)}=?$$
Use the Laplace transform to solve the given initial-value problem.
$$\displaystyle{y}{''}-{6}{y}'+{13}{y}={0},\ \ \ {y}{\left({0}\right)}={0},\ \ \ {y}'{\left({0}\right)}=-{9}$$
Use the Laplace transform to solve the given initial-value problem
$${y}{''}+{2}{y}'+{y}={0},{y}{\left({0}\right)}={1},{y}'{\left({0}\right)}={1}$$
Use Laplace transform to solve the following initial-value problem
$$y"+2y'+y=0$$
$$y(0)=1, y'(0)=1$$
a) \displaystyle{e}^{{-{t}}}+{t}{e}^{{-{t}}}\)
b) \displaystyle{e}^{t}+{2}{t}{e}^{t}\)
c) \displaystyle{e}^{{-{t}}}+{2}{t}{e}^{t}\)
d) \displaystyle{e}^{{-{t}}}+{2}{t}{e}^{{-{t}}}\)
e) \displaystyle{2}{e}^{{-{t}}}+{2}{t}{e}^{{-{t}}}\)
f) Non of the above
Let x(t) be the solution of the initial-value problem
(a) Find the Laplace transform F(s) of the forcing f(t).
(b) Find the Laplace transform X(s) of the solution x(t).
$$x"+8x'+20x=f(t)$$
$$x(0)=-3$$
$$x'(0)=5$$
$$\text{where the forcing } f(t) \text{ is given by }$$
$$f(t) = \begin{cases} t^2 & \quad \text{for } 0\leq t<2 ,\\ 4e^{2-t} & \quad \text{for } 2\leq t < \infty . \end{cases}$$
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:
$$\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}+{4}{\int_{{0}}^{{t}}}{y}{\left({t}-{w}\right)}{e}^{{-{4}{w}}}{d}{w}={3},{y}{\left({0}\right)}={0}$$
$${Y}{\left({s}\right)}={L}{\left\lbrace{y}{\left({t}\right)}\right)}{\rbrace}-?$$
$$2y"+4y'+17y=3\cos(2t)$$
$$y(0)=y'(0)=0$$
a)take Laplace transform of both sides of the given differntial equation to create corresponding algebraic equation and then solve for $$L\left\{y(t)\right\}$$ b) Express the solution $$y(t)$$ in terms of a convolution integral