Question

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

13.data-custom-editor="chemistry" $f\left(t\right)=t{e}^{4t}$

Short answer

The Laplace transform of above function is,

$L\left\{f\left(t\right)\right\}=\frac{1}{{\left(4-s\right)}^2}$

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 $f\left(t\right)=t.e^{4t}$

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.

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

$={\int }_0^{\infty }{e}^{-st}\left(t.e^{4t}\right)dt$

$={\int }_0^{\infty }t.e^{4t-st}dt$

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,

$I={\int }_0^{\infty }t.e^{4t-st}dt$

$u=t;v=e^{4t-st}$

$I=t\int e^{4t-st}dt-\int \left[\frac{d}{dt}t.\int e^{4t-st}dt\right]dt$

$I=t\left(\frac{e^{4t-st}}{4-s}\right)-\int \left[1.\left(\frac{e^{4t-st}}{4-s}\right)\right]dt$

$=t{\left(\frac{e^{4t-st}}{4-s}\right)}_0^{\infty }-\frac{1}{4-s}{\int }_0^{\infty }{e}^{4t-st}dt$

Simplification

$I=t{\left(\frac{e^{4t-st}}{4-s}\right)}_0^{\infty }-\frac{1}{4-s}{\left(\frac{e^{4t-st}}{4-s}\right)}_0^{\infty }$

$I=\left(\infty .\frac{e^{\left(4-s\right).\infty }}{4-s}-0.\frac{e^{\left(4-s\right).0}}{4-s}\right)-\frac{1}{4-s}{\left(\frac{e^{\left(4-s\right).\infty }}{4-s}-\frac{e^{\left(4-s\right).0}}{4-s}\right)}_0^{\infty }$

$I=0-\frac{1}{4-s}\left(0-\frac{e^0}{4-s}\right)$

$I=\frac{1}{{\left(4-s\right)}^2}$

Therefore the required Laplace transform of function is,

$L\left\{f\left(t\right)\right\}=\frac{1}{{\left(4-s\right)}^2}$