Question

# 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

Laplace transform
If the Laplace Transforms of fimetions $$y_1(t)=\int_0^\infty e^{-st}t^3dt , y_2(t)=\int_0^\infty e^{-st} \sin 2tdt , y_3(t)=\int_0^\infty e^{-st}e^t t^2dt$$ exist , then Which of the following is the value of $$L\left\{y_1(t)+y_2(t)+y_3(t)\right\}$$
$$a) \frac{3}{(s^4)}+\frac{2}{(s^2+4)}+\frac{2}{(s-1)^3}$$
$$B) \frac{(3!)}{(s^3)}+\frac{s}{(s^2+4)}+\frac{(2!)}{(s-1)^3}$$
$$c) \frac{3!}{(s^4)}+\frac{2}{(s^2+2)}+\frac{1}{(s-1)^3}$$
$$d) \frac{3!}{(s^4)}+\frac{4}{(s^2+4)}+\frac{2}{(s^3)} \cdot \frac{1}{(s-1)}$$
$$e) \frac{3!}{(s^4)}+\frac{2}{(s^2+4)}+\frac{2}{(s-1)^3}$$

2021-01-28
Step 1
The Laplace transform of the a function f(t) is defined by the integral:
$$L(f,s)=\int_0^\infty e^{-st} f(t)dt$$
for those s where the integral converges.
Step 2
First, find the Laplace transform for the value of $$f(t) = t^3$$
$$L(f,s)=\int_0^\infty e^{-st} f(t)dt$$
$$f(t)=t^3$$
$$L(t^3,s)=\int_0^\infty e^{-st}t^3 dt$$
$$\left\{\text{Use Laplace transform table: } L\left\{t^n\right\}=\frac{(n!)}{s^{n+1}}\right\}$$
$$\text{or } L\left\{t^3\right\}=\frac{6}{s^4}=\frac{3!}{(s^4)}$$
$$\text{hence } , L\left\{y_1(t)\right\}=\frac{3!}{(s^4)}$$
Step 3 Find the Laplace transform for the value of $$f(t) = \sin 2t$$
$$L(f,s)=\int_0^\infty e^{-st} f(t) dt$$
$$f(t)=\sin 2t$$
$$L(\sin 2t,s)=\int_0^\infty e^{-st} \sin 2tdt$$
$$\left\{\text{Use Laplace transform table: } L\left\{\sin(at)\right\}=\frac{a}{(s^2+a^2)}\right\}$$
$$L\left\{\sin(2t)\right\}=\frac{2}{(s^2+2^2)}$$
$$=\frac{2}{(s^2+4)}$$
$$L\left\{y_2(t)\right\}=\frac{2}{(s^2+4)}$$
Step 4
The Laplace transform for the value of $$f(t) = e^tt^2$$
$$L(f,s)=\int_0^\infty e^{-st} f(t)dt$$
$$f(t)=e^tt^2$$
$$L(e^tt^2,s)=\int_0^\infty e^{-st} (e^t t^2)dt$$
$$\text{Apply transform rule: }$$
$$if L{f(t)}=F(s) \text{ then } L{t^k f(t)}= (-1)^k \frac{d}{ds^k} (F(s))$$
$$f(t)=e^t , k=2$$
$$L\left\{e^t\right\}: \frac{1}{(s-1)} \text{ and } \frac{(d^2)}{(ds^2)(\frac{1}{s-1})}=\frac{2}{(s-1)^3}$$
$$=(-1)^2 \frac{2}{(s-1)^3}$$
$$=\frac{2}{(s-1)^3}$$
$$L(y_3)=\frac{2}{(s-1)^3}$$
Step 5
Now, the value of $$L\left\{ y_1(x) + y_2(x) + y_3(x)\right\}:$$
$$L\left\{y_1(t)+y_2(t)+y_3(t)\right\}=L(y_1)+L(y_2)+L(y_3)$$
$$=\frac{3!}{(s^4)}+\frac{2}{(s^2+4)}+\frac{2}{(s-1)^3}$$ Hence, the correct option is e