To determine the solution of the initial value problem {y}{''}+{4}{y}=delta{left({t}-piright)}-delta{left({t}-{2}piright)},{y}{left({0}right)}={0},{y}'{left({0}right)}={0} and draw the graph of the solution.

FizeauV 2021-01-16 Answered
To determine the solution of the initial value problem
y+4y=δ(tπ)δ(t2π),y(0)=0,y(0)=0
and draw the graph of the solution.
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Expert Answer

Nola Robson
Answered 2021-01-17 Author has 94 answers

Decision:
Let the differential equation y+4y=δ(tπ)δ(t2π).
Applying the Laplace transform to the differential equation
L{y+4y}=L{δ(tπ)δ(t2π)}
L{y}+4L{y}=L{δ(tπ)}L{δ(t2π)}
By using L{fn(t)}=snF(s)sn1f(0)fn1(0) and
L{δ(tt0)}=est0,
[s2Y(s)sy(0)y(0)]+4Y(s)=eπse2πs
Using the initial conditions y(0)=0 and y(0)=0,
s2Y(s)+4Y(s)=eπse2πs
Y(s)[s2+4]=eπse2πs
Y(s)=eπse2πss2+4
Now, applying the inverse Laplace transformation on both sides,
L1{Y(s)}=L1{eπss2+4e2πss2+4}
L1{Y(s)}=12L1{2eπss2+22}12L1{2e2πss2+22}
By using L1{ecsG(s)}=uc(t)g(tc) and properties of inverse Laplace,

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