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
asked 2021-01-27
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 \(L\left\{y_1(t)+y_2(t)+y_3(t)\right\}\)
\(a) \frac{3}{(s^4)}+\frac{2}{(s^2+4)}+\frac{2}{(s-1)^3}\)
\(B) \frac{(3!)}{(s^3)}+\frac{s}{(s^2+4)}+\frac{(2!)}{(s-1)^3}\)
\(c) \frac{3!}{(s^4)}+\frac{2}{(s^2+2)}+\frac{1}{(s-1)^3}\)
\(d) \frac{3!}{(s^4)}+\frac{4}{(s^2+4)}+\frac{2}{(s^3)} \cdot \frac{1}{(s-1)}\)
\(e) \frac{3!}{(s^4)}+\frac{2}{(s^2+4)}+\frac{2}{(s-1)^3}\)

Answers (1)

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

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