To determine the solution of the initial value problem {y}{''}+{4}{y}= sin{{t}}+{u}_{pi}{left({t}right)} sin{{left({t}-piright)}}: y(0) = 0, y'(0) = 0. Also, draw the graphs of the solution and of the forcing function and explain the relation between the solution and the forcing function..

Anish Buchanan 2021-03-07 Answered
To determine the solution of the initial value problem
y+4y=sint+uπ(t)sin(tπ):
y(0)=0,
y(0)=0.
Also, draw the graphs of the solution and of the forcing function and explain the relation between the solution and the forcing function..
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Expert Answer

Viktor Wiley
Answered 2021-03-08 Author has 84 answers

Applying the Laplace transform to both sides of the differential equation
y+4y=sint+uπ(t)sin(tπ).
L{y+4y}=L{sint+uπ(t)sin(tπ)}
L{y}+L{4y}=L{sint}+L{uπ(t0sin(tπ)}
By using L{fn(t)}=snF(s)sn1(0)f(n1)(0) and
ecsG(s)=L{uc(t)g(tc)},L{sint}=1s2+1
[s2Y(s)sy(0)y(0)]+4Y(s)=1s2+1+eπss2+1
Using the initial conditions, y(0)=0 and y(0)=0,
s2Y(s)+4Y(s)=1+eπss2+1
Y(s)[s2+4]=1+eπss2+1
Y(s)=1+eπs(s2+1)(s2+4)
By using the partial fraction 1+eπs(s2+1)(s2+4)=13(1+eπs)1(s2+1)+1(s2+4)
Y(s)=13(1+eπs)1(s2+1)+1(s2+4)
Applying the inverse Laplace transformation on both sides,

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