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

permaneceerc 2021-01-27 Answered
If the Laplace Transforms of fimetions y1(t)=0estt3dt,y2(t)=0estsin2tdt,y3(t)=0estett2dt exist , then Which of the following is the value of L{y1(t)+y2(t)+y3(t)}
a)3(s4)+2(s2+4)+2(s1)3
B)(3!)(s3)+s(s2+4)+(2!)(s1)3
c)3!(s4)+2(s2+2)+1(s1)3
d)3!(s4)+4(s2+4)+2(s3)1(s1)
e)3!(s4)+2(s2+4)+2(s1)3
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Expert Answer

SabadisO
Answered 2021-01-28 Author has 108 answers
Step 1
The Laplace transform of the a function f(t) is defined by the integral:
L(f,s)=0estf(t)dt
for those s where the integral converges.
Step 2
First, find the Laplace transform for the value of f(t)=t3
L(f,s)=0estf(t)dt
f(t)=t3
L(t3,s)=0estt3dt
{Use Laplace transform table: L{tn}=(n!)sn+1}
or L{t3}=6s4=3!(s4)
hence ,L{y1(t)}=3!(s4)
Step 3 Find the Laplace transform for the value of f(t)=sin2t
L(f,s)=0estf(t)dt
f(t)=sin2t
L(sin2t,s)=0estsin2tdt
{Use Laplace transform table: L{sin(at)}=a(s2+a2)}
L{sin(2t)}=2(s2+22)
=2(s2+4)
L{y2(t)}=2(s2+4)
Step 4
The Laplace transform for the value of f(t)=ett2
L(f,s)=0estf(t)dt
f(t)=ett2
L(ett2,s)=0est(ett2)dt
Apply transform rule: 
ifLf(t)=F(s) then Ltkf(t)=(1)kddsk(F(s))
f(t)=et,k=2
L{et}:1(s1) and (d2)(ds2)(1s1)=2(s1)3
=(1)22(s1)3
=2(s1)3
L(y3)=2(s1)3
Step 5
Now, the value of L{y1(x)+y2(x)+y3(x)}:
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