If the Laplace Transforms of fimetions y1(t)=∫0∞e−stt3dt,y2(t)=∫0∞e−stsin⁡2tdt,y3(t)=∫0∞e−stett2dt exist , then Which of the following is the value of L{y1(t)+y2(t)+y3(t)} a)3(s4)+2(s2+4)+2(s−1)3 B)(3!)(s3)+s(s2+4)+(2!)(s−1)3 c)3!(s4)+2(s2+2)+1(s−1)3 d)3!(s4)+4(s2+4)+2(s3)⋅1(s−1) e)3!(s4)+2(s2+4)+2(s−1)3

Name: If the Laplace Transforms of fimetions y1(t)=∫0∞e−stt3dt,y2(t)=∫0∞e−stsin⁡2tdt,y3(t)=∫0∞e−stett2dt exist , then Which of the following is the value of L{y1(t)+y2(t)+y3(t)} a)3(s4)+2(s2+4)+2(s−1)3 B)(3!)(s3)+s(s2+4)+(2!)(s−1)3 c)3!(s4)+2(s2+2)+1(s−1)3 d)3!(s4)+4(s2+4)+2(s3)⋅1(s−1) e)3!(s4)+2(s2+4)+2(s−1)3
Uploaded: 2022-02-23T17:22:57+00:00
Description: If the Laplace Transforms of fimetions y1(t)=∫0∞e−stt3dt,y2(t)=∫0∞e−stsin⁡2tdt,y3(t)=∫0∞e−stett2dt exist , then Which of the following is the value of L{y1(t)+y2(t)+y3(t)} a)3(s4)+2(s2+4)+2(s−1)3 B)(3!)(s3)+s(s2+4)+(2!)(s−1)3 c)3!(s4)+2(s2+2)+1(s−1)3 d)3!(s4)+4(s2+4)+2(s3)⋅1(s−1) e)3!(s4)+2(s2+4)+2(s−1)3

Question

If the Laplace Transforms of fimetions y1(t)=∫0∞e−stt3dt,y2(t)=∫0∞e−stsin⁡2tdt,y3(t)=∫0∞e−stett2dt exist , then Which of the following is the value of L{y1(t)+y2(t)+y3(t)}  a)3(s4)+2(s2+4)+2(s−1)3  B)(3!)(s3)+s(s2+4)+(2!)(s−1)3  c)3!(s4)+2(s2+2)+1(s−1)3  d)3!(s4)+4(s2+4)+2(s3)⋅1(s−1)  e)3!(s4)+2(s2+4)+2(s−1)3

SabadisO · Accepted Answer

Step 1  The Laplace transform of the a function f(t) is defined by the integral:  L(f,s)=∫0∞e−stf(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)=∫0∞e−stf(t)dt  f(t)=t3  L(t3,s)=∫0∞e−stt3dt  {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)=sin⁡2t  L(f,s)=∫0∞e−stf(t)dt  f(t)=sin⁡2t  L(sin⁡2t,s)=∫0∞e−stsin⁡2tdt  {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)=∫0∞e−stf(t)dt  f(t)=ett2  L(ett2,s)=∫0∞e−st(ett2)dt  Apply transform rule:   ifLf(t)=F(s) then Ltkf(t)=(−1)kddsk(F(s))  f(t)=et,k=2  L{et}:1(s−1) and (d2)(ds2)(1s−1)=2(s−1)3  =(−1)22(s−1)3  =2(s−1)3  L(y3)=2(s−1)3  Step 5  Now, the value of L{y1(x)+y2(x)+y3(x)}:

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

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