Question

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

12.$f\left(t\right)=e^{-2t-5}$.

Short answer

The Laplace transform of above function is,

$L\left\{f\left(t\right)\right\}=\frac{e^{-5}}{2+s}$

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$f\left(t\right)=e^{-2t-5}$

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.

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

$={\int }_0^{\infty }{e}^{-st}\left(e^{-2t-5}\right)dt$

$={\int }_0^{\infty }{e}^{-2t-5-st}dt$

Simplification

$I={\int }_0^{\infty }{e}^{-\left(2+s\right)t-5}dt$

$={\left\{\frac{-e^{-\left(2+s\right)t-5}}{2+s}\right\}}_0^{\infty }$

$I=-\frac{e^{-\left(2+s\right).\infty -5}}{2+s}+\frac{e^{-\left(2+s\right).0-5}}{2+s}$

$I=0+\frac{e^{-5}}{2+s}$

$I=\frac{e^{-5}}{2+s}$

Therefore the required Laplace transform of function is,

$L\left\{f\left(t\right)\right\}=\frac{e^{-5}}{2+s}$