# If the Laplace Transforms of fimetions y_1(t)=int_0^infty e^{-st}t^3dt , y_2(t)=int_0^infty e^{-st} sin 2tdt , y_3(t)=int_0^infty e^{-st}e^t t^2dt exi

If the Laplace Transforms of fimetions ${y}_{1}\left(t\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}{t}^{3}dt,{y}_{2}\left(t\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}\mathrm{sin}2tdt,{y}_{3}\left(t\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}{e}^{t}{t}^{2}dt$ exist , then Which of the following is the value of $L\left\{{y}_{1}\left(t\right)+{y}_{2}\left(t\right)+{y}_{3}\left(t\right)\right\}$
$a\right)\frac{3}{\left({s}^{4}\right)}+\frac{2}{\left({s}^{2}+4\right)}+\frac{2}{\left(s-1{\right)}^{3}}$
$B\right)\frac{\left(3!\right)}{\left({s}^{3}\right)}+\frac{s}{\left({s}^{2}+4\right)}+\frac{\left(2!\right)}{\left(s-1{\right)}^{3}}$
$c\right)\frac{3!}{\left({s}^{4}\right)}+\frac{2}{\left({s}^{2}+2\right)}+\frac{1}{\left(s-1{\right)}^{3}}$
$d\right)\frac{3!}{\left({s}^{4}\right)}+\frac{4}{\left({s}^{2}+4\right)}+\frac{2}{\left({s}^{3}\right)}\cdot \frac{1}{\left(s-1\right)}$
$e\right)\frac{3!}{\left({s}^{4}\right)}+\frac{2}{\left({s}^{2}+4\right)}+\frac{2}{\left(s-1{\right)}^{3}}$
You can still ask an expert for help

## Want to know more about Laplace transform?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Step 1
The Laplace transform of the a function f(t) is defined by the integral:
$L\left(f,s\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt$
for those s where the integral converges.
Step 2
First, find the Laplace transform for the value of $f\left(t\right)={t}^{3}$
$L\left(f,s\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt$
$f\left(t\right)={t}^{3}$
$L\left({t}^{3},s\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}{t}^{3}dt$

Step 3 Find the Laplace transform for the value of $f\left(t\right)=\mathrm{sin}2t$
$L\left(f,s\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt$
$f\left(t\right)=\mathrm{sin}2t$
$L\left(\mathrm{sin}2t,s\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}\mathrm{sin}2tdt$

$L\left\{\mathrm{sin}\left(2t\right)\right\}=\frac{2}{\left({s}^{2}+{2}^{2}\right)}$
$=\frac{2}{\left({s}^{2}+4\right)}$
$L\left\{{y}_{2}\left(t\right)\right\}=\frac{2}{\left({s}^{2}+4\right)}$
Step 4
The Laplace transform for the value of $f\left(t\right)={e}^{t}{t}^{2}$
$L\left(f,s\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt$
$f\left(t\right)={e}^{t}{t}^{2}$
$L\left({e}^{t}{t}^{2},s\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}\left({e}^{t}{t}^{2}\right)dt$

$f\left(t\right)={e}^{t},k=2$

$=\left(-1{\right)}^{2}\frac{2}{\left(s-1{\right)}^{3}}$
$=\frac{2}{\left(s-1{\right)}^{3}}$
$L\left({y}_{3}\right)=\frac{2}{\left(s-1{\right)}^{3}}$
Step 5
Now, the value of $L\left\{{y}_{1}\left(x\right)+{y}_{2}\left(x\right)+{y}_{3}\left(x\right)\right\}:$