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

Step1
Given
$(s)=2t^2+{\int }_0^t\sin [2(t-\tau )]x(\tau )d\tau$
Step2
Taking Laplace transform on both sides of the equation
$L\{x(t)\}=L\{2t^2+{\int }_0^t\sin [2(t-\tau )]x(\tau )d\tau \}$
By using linearity property of Laplace transform
$L[af(t)+bg(t)]=aL\{f(t)\}+bL\{f(t)\}$
where a, b are constants. $L\{x(t)\}=2L\left\{t^2\right\}+L\{{\int }_0^{t}sin[2(t-\tau )]x(\tau )d\tau \}$
Step3
By using convolution theorem
$L\{{\int }_0^t\sin [2(t-\tau )]x(\tau )d\tau \}=L\{\sin (2t)\}L\{x(t)\}$
Therefore,
$L\{x(t)\}=2L\left\{t^2\right\}+L\{\sin (2t)\}L\{x(t)\}$
Use the formula:
$L\left\{t^n\right\}=\frac{n!}{s^{n+1}},L\{\sin at\}=\frac{a}{s^2+a^2}$
It gives
$x(s)=2\left(\frac{2!}{s^{2+1}}\right)+\left(\frac{2}{s^2+2^2}\right)x(s)$
$x(s)=\frac{4}{s^3}+\frac{2}{s^2+4}x(s)$
Solve above equation for x(s)
$x(s)-\frac{2}{s^2+4}x(s)=\frac{4}{s^3}$
$(1-\frac{2}{s^2+4})x(s)=\frac{4}{s^3}$
$\left(\frac{s^2+2}{s^2+4}\right)x(s)=\frac{4}{s^3}$
Therefore,
$x(s)=\frac{4(s^2+4)}{s^3(s^2+2)}$
Step 4
By using partial fractions:
$\frac{4(s^2+4)}{s^3(s^2+2)}=\frac{-2}{s}+\frac{8}{s^3}+\frac{2s}{s^2+2}$
Taking inverse Laplace transform on both the sides.
$L^{-1}\left\{\frac{4(s^2+4)}{s^3(s^2+2)}\right\}=L^{-1}\{\frac{-2}{s}+\frac{8}{s^3}+\frac{2s}{s^2+2}\}$
Using linearity property of Inverse Laplace transform,
$L^{-1}\{x(s)\}=-2L^{-1}\left\{\frac{1}{s}\right\}+8L^{-1}\left\{\frac{1}{s^3}\right\}+2L^{-1}\left\{\frac{s}{s^2+2}\right\}$
Step 5
Use formulae for Inverse Laplace transform,