Use the Laplace transform to solve the given integral equation. x(t)=2t2+∫0tsin⁡[2(t−τ)]x(τ)dτ

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Step 1 Taking Laplace transform on both sides of the equation L{x(t)}=L{2t2+∫0tsin⁡[2(t−τ)]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(t−τ)]x(τ)dτ} Step 2 By using convolution theorem L{∫0tsin⁡[2(t−τ)]x(τ)dτ}=L{sin⁡(2t)}L{x(t)} Use the formulae: L{tn}=n!sn+1, L{sin⁡at}=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   (1−2s2+4)x(s)=4s3 (s2+2s2+4)x(s)=4s3 Therefore, x(s)=4(s2+4)s3(s2+2)  Step 3 By using partial fractions: 4(s2+4)s3(s2+2)=−2s+8s3+2s(s2+2) Taking inverse Laplace transform on both the sides.

Use the Laplace transform to solve the given integral equation.

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