Question

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

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

Short answer

The Laplace transform of above function is,

$L\left\{e^{-t}.\sin t\right\}=\frac{1}{\left(4s^2+4s+1\right)+1};s>-\frac{1}{2}$

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

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.

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

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

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

Tosimplify it we use a direct formula;

So formula is$L\left\{e^{at}.\sin bt\right\}=\frac{b}{{\left(s-a\right)}^2+b^2};s>a$

Thereforeoncomparing,

$a=-\left(1+s\right);b=1$

Applying the above formula;

$L\left\{e^{-\left(1+s\right)t}.\sin t\right\}=\frac{1}{{\left(s+1+s\right)}^2+1^2};s>-(1+s)$

Simplification

$L\left\{e^{-\left(1+s\right)t}.\sin t\right\}=\frac{1}{{\left(2s+1\right)}^2+1^2};s>-1-s$

$L\left\{e^{-\left(1+s\right)t}.\sin t\right\}=\frac{1}{\left(4s^2+4s+1\right)+1};s+s>-1$

$L\left\{e^{-\left(1+s\right)t}.\sin t\right\}=\frac{1}{\left(4s^2+4s+1\right)+1};s+s>-1$

$L\left\{e^{-\left(1+s\right)t}.\sin t\right\}=\frac{1}{\left(4s^2+4s+1\right)+1};s>-\frac{1}{2}$

Therefore the required Laplace transform of function is,

$L\left\{e^{-t}.\sin t\right\}=\frac{1}{\left(4s^2+4s+1\right)+1};s>-\frac{1}{2}$