Use the Laplace transform to solve the given integral equation (s)=2t^2+int_0^tsinleft[2(t-tau)right]x(tau)d tau

Dolly Robinson 2020-11-02 Answered
Use the Laplace transform to solve the given integral equation
(s)=2t2+0tsin[2(tτ)]x(τ)dτ
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Expert Answer

Corben Pittman
Answered 2020-11-03 Author has 83 answers
Step1
Given
(s)=2t2+0tsin[2(tτ)]x(τ)dτ
Step2
Taking Laplace transform on both sides of the equation
L{x(t)}=L{2t2+0tsin[2(tτ)]x(τ)dτ}
By using linearity property of Laplace transform
L[af(t)+bg(t)]=aL{f(t)}+bL{f(t)}
where a, b are constants. L{x(t)}=2L{t2}+L{0tsin[2(tτ)]x(τ)dτ}
Step3
By using convolution theorem
L{0tsin[2(tτ)]x(τ)dτ}=L{sin(2t)}L{x(t)}
Therefore,
L{x(t)}=2L{t2}+L{sin(2t)}L{x(t)}
Use the formula:
L{tn}=n!sn+1,L{sinat}=as2+a2
It gives
x(s)=2(2!s2+1)+(2s2+22)x(s)
x(s)=4s3+2s2+4x(s)
Solve above equation for x(s)
x(s)2s2+4x(s)=4s3
(12s2+4)x(s)=4s3
(s2+2s2+4)x(s)=4s3
Therefore,
x(s)=4(s2+4)s3(s2+2)
Step 4
By using partial fractions:
4(s2+4)s3(s2+2)=2s+8s3+2ss2+2
Taking inverse Laplace transform on both the sides.
L1{4(s2+4)s3(s2+2)}=L1{2s+8s3+2ss2+2}
Using linearity property of Inverse Laplace transform,
L1{x(s)}=2L1{1s}+8L1{1s3}+2L1{ss2+2}
Step 5
Use formulae for Inverse Laplace transform,
L1{1s}=1,L1{1sn}=tn1(n1)!,L1{ss2+a2
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