# Let f(t) be a function on displaystyle{left[{0},inftyright)}. 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)}}={leftlbracebegin{matrix}{1}-{t}&{0}<{t}<{1}{0}&{1}<{t}end{matrix}right.}

Question
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.
$$\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{1}-{t}&{0}<{t}<{1}\\{0}&{1}<{t}\end{matrix}\right.}$$</span>

2020-12-28
Step 1
We know that , $$\displaystyle{F}{\left({s}\right)}={\int_{{0}}^{\infty}}{e}^{{-{s}{t}}} f{{\left({t}\right)}}{\left.{d}{t}\right.}$$
Here $$\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{1}-{t}&{0}<{t}<{1}\\{0}&{1}<{t}\end{matrix}\right.}$$</span>
from $$\displaystyle{F}{\left({s}\right)}={\int_{{0}}^{\infty}}{\left( f{{\left({t}\right)}}\right)}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}={\int_{{0}}^{{1}}}{\left({1}-{t}\right)}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}+{\int_{{1}}^{\infty}}{0}\cdot{e}^{{-{s}{t}}}{\left.{d}{t}\right.}$$
$$\displaystyle={\int_{{0}}^{{1}}}{\left({1}-{t}\right)}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}+{0}$$
$$\displaystyle{F}{\left({s}\right)}={{\left[\frac{{{\left({1}-{t}\right)}{e}^{{-{s}{t}}}}}{{-{s}}}-\frac{{{\left(-{1}\right)}{e}^{{-{s}{t}}}}}{{{s}^{2}}}\right]}_{{0}}^{{1}}}$$
$$\displaystyle{F}{\left({s}\right)}=\frac{1}{{s}}+\frac{{{e}^{{-{s}}}}}{{s}}-\frac{1}{{s}^{2}}$$
Step 2
This is required Laplace Transform.

### Relevant Questions

$$f(t)=3e^{2t}$$
Determine L[f]
Let f be a function defined on an interval $$[0,\infty)$$
The Laplace transform of f is the function F(s) defined by
$$F(s) =\int_0^\infty e^{-st}f(t)dt$$
provided that the improper integral converges. We will usually denote the Laplace transform of f by L[f].
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) - ?
use the Laplace transform to solve the given initial-value problem.
$$y"+y=f(t)$$
$$y(0)=0 , y'(0)=1$$ where
$$\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{1}&{0}\le{t}<\frac{\pi}{{2}}\\{0}&{t}\ge\frac{\pi}{{2}}\end{matrix}\right.}$$
$$\text{Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by }$$
$$F(s)=\int_0^\infty e^{-st} f(t)dt \(\text{where we assume s is a positive real number. For example, to find the Laplace transform of } f(t)=e^{-t} \text{ , the following improper integral is evaluated using integration by parts:} \(F(s)=\int_0^\infty e^{-st}e^{-t}dt=\int_0^\infty e^{-(s+1)t}dt=\frac{1}{s+1}$$
$$\text{ Verify the following Laplace transforms, where u is a real number. }$$
$$f(t)=t \rightarrow F(s)=\frac{1}{s^2}$$
Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by
$$F(s)=\int_0^\infty e^{-st}f(t)dt$$
where we assume s is a positive real number. For example, to find the Laplace transform of f(t) = e^{-t}, the following improper integral is evaluated using integration by parts:
$$F(s)=\int_0^\infty e^{-st}e^{-t}dt=\int_0^\infty e^{-(s+1)t}dt=\frac{1}{(s+1)}$$
Verify the following Laplace transforms, where u is a real number.
$$f(t)=1 \rightarrow F(s)=\frac{1}{s}$$
Solve for Y(s), the Laplace transform of the solution y(t) to the initial value problem below.
$$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.}$$
Y(s)-?
Let x(t) be the solution of the initial-value problem
(a) Find the Laplace transform F(s) of the forcing f(t).
(b) Find the Laplace transform X(s) of the solution x(t).
$$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}$$
Solve the initial value problem $$\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.}$$ using the Laplace transform.
$$\begin{cases}t & 0\leq t<1\\ e^t & t\geq1 \end{cases}$$
$$L(f(t))=\int_0^1te^{-st}dt+\int_1^\infty e^{-(s+1)t}dt$$
$${L}{\left\lbrace{e}^{{-{t}}}\cdot{e}^{t} \cos{{\left({t}\right)}}\right\rbrace}$$