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# 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)}}={leftlbracebegin{matrix}{t}&{t}<{4}{5}&{t}>{4}end{matrix}right.} Y(s)-?

Question
Laplace transform
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)-?

2020-11-17
Step 1 Basics
Here we use Second Shifting Property and Laplace Transform of Derivative as follows
Step 1
Laplace transform of derivative
$${L}{\left\lbrace{f}\text{}{\left({t}\right)}\right\rbrace}={s}^{2}{L}{\left\lbrace{f}\right\rbrace}-{\mathsf{{{0}}}}-{f}'{\left({0}\right)}$$
second shifting property
$${L}{\left\lbrace f{{\left({t}-{t}_{{0}}\right)}}\right\rbrace}={e}^{{-{s}{t}_{{0}}}}{L}{\left\lbrace f{{\left({t}\right)}}\right\rbrace}$$
Step 2 Explanation
$$y"+y=g(t) , y(0)=-4 , y'(0)=0$$
where $$g{{\left({t}\right)}}={\left\lbrace\begin{matrix}{t}&{t}<{4}\\{5}&{t}>{4}\end{matrix}\right.}$$
Taking Laplace transform both side of above equation , we get
$${L}{\left\lbrace{y}\text{}\right\rbrace}+{L}{\left\lbrace{y}\right\rbrace}={L}{\left\lbrace g{{\left({t}\right)}}\right\rbrace}$$
$${\left[{s}^{2}{L}{\left\lbrace{y}{\left({t}\right)}\right\rbrace}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}\right]}+{L}{\left\lbrace{y}\right\rbrace}={L}{\left\lbrace{t}{u}{\left({t}\right)}+{5}{u}{\left({t}-{u}\right)}\right]}$$
where u(t) is the step function
$${\left[\text{use }\ {L}{\left\lbrace{t}^{n}{u}{\left({n}\right)}\right\rbrace}=\frac{{{n}!}}{{{s}^{{{n}+{1}}}}}\right]}$$
shifting property as $${L}{\left\lbrace f{{\left({t}-{t}_{{0}}\right)}}\right\rbrace}={e}^{{-{s}{t}_{{0}}}}{L}{\left\lbrace f{{\left({t}\right)}}\right\rbrace}$$
$${s}^{2}{L}{\left\lbrace{y}\right\rbrace}+{4}{s}+{L}{\left\lbrace{y}\right\rbrace}=\frac{1}{{s}^{2}}+{e}^{{-{4}{s}}}{\left(\frac{5}{{s}}\right)}$$
$${L}{\left\lbrace{y}\right\rbrace}{\left[{s}^{2}+{1}\right]}=\frac{1}{{s}^{2}}+\frac{{{5}{e}^{{-{4}{s}}}}}{{s}}-{4}{s}$$
$${L}{\left\lbrace{y}\right\rbrace}=\frac{1}{{{s}^{2}{\left({s}^{2}+{1}\right)}}}+\frac{{{5}{e}^{{-{4}{s}}}}}{{{s}{\left({s}^{2}+{1}\right)}}}-{4}\frac{s}{{{s}^{2}+{1}}}$$
$${y}{\left({s}\right)}=\frac{1}{{{s}^{2}{\left({s}^{2}+{1}\right)}}}+\frac{{{5}{e}^{{-{4}{s}}}}}{{{s}{\left({s}^{2}+{1}\right)}}}-{4}\frac{s}{{{s}^{2}+{1}}}$$

### Relevant Questions

use the Laplace transform to solve the given initial-value problem.
$$y"+y=f(t)$$
$$y(0)=0 , y'(0)=1$$ where
$$\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{1}&{0}\le{t}<\frac{\pi}{{2}}\\{0}&{t}\ge\frac{\pi}{{2}}\end{matrix}\right.}$$
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 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:
$$\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}+{4}{\int_{{0}}^{{t}}}{y}{\left({t}-{w}\right)}{e}^{{-{4}{w}}}{d}{w}={3},{y}{\left({0}\right)}={0}$$
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) - ?
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 following initial value problem:
$$2y"+4y'+17y=3\cos(2t)$$
$$y(0)=y'(0)=0$$
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
Use the Laplace transform to solve the given initial-value problem.
$$\displaystyle{y}{''}+{8}{y}'+{41}{y}=\delta{\left({t}-\pi\right)}+\delta{\left({t}-{3}\pi\right)},\ \ \ \ {y}{\left({0}\right)}={1},\ \ \ \ {y}'{\left({0}\right)}={0}\ \ {y}{\left({t}\right)}=?$$
$${y}\text{}{\left({t}\right)}-{y}'{\left({t}\right)}={e}^{t} \cos{{\left({t}\right)}}+ \sin{{\left({t}\right)}},{y}{\left({0}\right)}={0},{y}'{\left({0}\right)}={0}$$
For what value of the constant c is the function f continuous on $$\displaystyle{\left(−∞,+∞\right)}?$$
 $$\displaystyle{f{{\left({x}\right)}}}={\left\lbrace{\left({c}{x}^{{2}}\right)}+{4}{x},{\left({x}^{{3}}\right)}-{c}{x}\right\rbrace}{\quad\text{if}\quad}{x}{<}{5},{\quad\text{if}\quad}{x}\Rightarrow{5}$$ c=
$$y'-y=t e^t \sin t$$
$$y(0)=0$$