Question

In Problems 1–18 use Definition 7.1.1 to find$L\left\{f\left(t\right)\right\}.$

14.$f\left(t\right)=t^2.e^{-2t}$

Short answer

The Laplace transform of above function is,

$L\left\{f\left(t\right)\right\}=-\frac{2}{{\left(2+s\right)}^3}$

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$f\left(t\right)=t^2.e^{-2t}$

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 as$[0,\infty ),$ Laplacian if f is$L\left\{f\left(t\right)\right\}$expressed as the integrals.

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

$={\int }_0^{\infty }{e}^{-st}\left(t^2.e^{-2t}\right)dt$

$={\int }_0^{\infty }{t}^2.e^{-2t-st}dt$

Letsolve it first, by using integral by parts formula;

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,

$I={\int }_0^{\infty }{t}^2.e^{-2t-st}dt$

$u=t;v=e^{-2t-st}$

$I=t^2\int e^{-2t-st}dt-\int \left[\frac{d}{dt}{t}^2.\int e^{-2t-st}dt\right]dt$

$={\left\{t^2\left(-\frac{e^{-\left(2+s\right)t}}{2+s}\right)\right\}}_0^{\infty }-\int \left[2t.\left(-\frac{e^{-\left(2+s\right)t}}{2+s}\right)\right]dt$

$={\left\{t^2\left(-\frac{e^{-\left(2+s\right)t}}{2+s}\right)\right\}}_0^{\infty }+\frac{2}{2+s}\int t.e^{-\left(2+s\right)t}dt$

$={\left\{t^2\left(-\frac{e^{-\left(2+s\right)t}}{2+s}\right)\right\}}_0^{\infty }+\frac{2}{2+s}{\left\{t\int e^{-2t-st}dt-\int \left[\frac{d}{dt}t.\int e^{-2t-st}dt\right]dt\right\}}_0^{\infty }$

$={\left\{t^2\left(-\frac{e^{-\left(2+s\right)t}}{2+s}\right)\right\}}_0^{\infty }+\frac{2}{2+s}{\left\{t\left(-\frac{e^{-2t-st}}{2+s}\right)-\int \left[1.\left(-\frac{e^{-2t-st}}{2+s}\right)\right]dt\right\}}_0^{\infty }$

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

Simplification

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

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

$=-\frac{2}{{\left(2+s\right)}^3}$

$I=-\frac{2}{{\left(2+s\right)}^3}$

Therefore the required Laplace transform of function is,

$L\left\{f\left(t\right)\right\}=-\frac{2}{{\left(2+s\right)}^3}$