Question

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

11.$f\left(t\right)=e^{t+7}$

Short answer

The Laplace transform of above function is,

$L\left\{f\left(t\right)\right\}=-\frac{e^7}{1-s}$

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$f\left(t\right)=e^{t+7}$

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.

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

$={\int }_0^{\infty }{e}^{-st}\left(e^{t+7}\right)dt$

$={\int }_0^{\infty }{e}^{t+7-st}dt$

Simplification

$I={\int }_0^{\infty }{e}^{\left(1-s\right)t+7}dt$

$={\left\{\frac{e^{\left(1-s\right)t+7}}{1-s}\right\}}_0^{\infty }$

$I=\frac{e^{\left(1-s\right).\infty +7}}{1-s}-\frac{e^{\left(1-s\right).0+7}}{1-s}$

$I=\frac{0}{1-s}-\frac{e^7}{1-s}$

$I=-\frac{e^7}{1-s}$

Therefore the required Laplace transform of function is,

$L\left\{f\left(t\right)\right\}=-\frac{e^7}{1-s}$