Question

In Problems 1–18 use Definition 7.1.1 to find

9.

Short answer

The Laplace transform of above function is,

$I=\frac{1}{s^2}{e}^{-s}+\frac{1}{s}-\frac{1}{s^2}$

Step-by-step solution

Definition 7.1.1 Laplace transform

Let f be a function define for$t⩾0$.Then the integral

$L\left\{f\left(t\right)\right\}={\int }_0^{\infty }{e}^{-st}f\left(t\right)dt$

is said to be Laplace transform of f provide that integral converges.

Applying the definition

Consider the function $f\left(t\right)=\left\{\begin{array}{c}1-t,0<t<1\\ 0,t>1\end{array}\right.$

The objective is to find $L\left\{f\left(t\right)\right\}$using the definition.

Note that, the function f is defined for$t⩾0$.

From the definition,

$L\left\{f\left(t\right)\right\}={\int }_0^{\infty }{e}^{-st}f\left(t\right)dt$

Since f is defined in two pieces$[0,1)$ and $[1,\infty )$Laplacian if f is$L\left\{f\left(t\right)\right\}$expressed as the sum of two integrals.

$L\left\{f\left(t\right)\right\}={\int }_0^1{e}^{-st}f\left(t\right)dt+{\int }_1^{\infty }{e}^{-st}f\left(t\right)dt$

$={\int }_0^1{e}^{-st}\left(1-t\right)dt+{\int }_1^{\infty }{e}^{-st}\left(0\right)dt$

$={\int }_0^1\left(1-t\right).e^{-st}dt+0$

Let solve first, by using integral by parts formula

$I={\int }_0^1\left(1-t\right).e^{-st}dt$

Soformula is,$I=\int u.vdt=u\int vdt-\int \left[\frac{d}{dt}u.\int vdt\right]dt$

Where u and v we choose according to ILATE rule;

I=Inverse

L=Logarithmic

A=Arithmetic

T=Trigonometry

E=Exponential

Asarithmeticfunctioncomesfirst,

Therefore,

$I={\int }_0^1\left(1-t\right).e^{-st}dt$

$u=1-t;v=e^{-st}$

$=\left\{\left(1-t\right)\int e^{-st}dt-\int \left[\frac{d}{dt}\left(1-t\right).\int e^{-st}dt\right]dt\right\}$

$=\left\{\left(1-t\right)\left(\frac{-e^{-st}}{s}\right)-\int \left[-1.\left(\frac{-e^{-st}}{s}\right)\right]dt\right\}$

$I={\left\{\left(1-t\right)\left(-\frac{e^{-st}}{s}\right)-\frac{1}{s^2}{e}^{-st}\right\}}_0^1$

Simplification

$I={\left\{\left(1-t\right)\left(-\frac{e^{-st}}{s}\right)-\frac{1}{s^2}{e}^{-st}\right\}}_0^1$

$I=\left(\left(1-1\right).\left(-\frac{e^{-s.1}}{s}\right)-\frac{1}{s^2}{e}^{-s.1}\right)-\left(\left(1-0\right).\left(-\frac{e^{-s.0}}{s}\right)-\frac{1}{s^2}{e}^{-s.0}\right)$

$I=\frac{1}{s^2}{e}^{-s}+\frac{1}{s}-\frac{1}{s^2}$

Therefore the required Laplace transform of function is,

$I=\frac{1}{s^2}{e}^{-s}+\frac{1}{s}-\frac{1}{s^2}$