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

Use the Laplace transform to solve the given integral equation
$\left(s\right)=2{t}^{2}+{\int }_{0}^{t}\mathrm{sin}\left[2\left(t-\tau \right)\right]x\left(\tau \right)d\tau$
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Corben Pittman
Step1
Given
$\left(s\right)=2{t}^{2}+{\int }_{0}^{t}\mathrm{sin}\left[2\left(t-\tau \right)\right]x\left(\tau \right)d\tau$
Step2
Taking Laplace transform on both sides of the equation
$L\left\{x\left(t\right)\right\}=L\left\{2{t}^{2}+{\int }_{0}^{t}\mathrm{sin}\left[2\left(t-\tau \right)\right]x\left(\tau \right)d\tau \right\}$
By using linearity property of Laplace transform
$L\left[af\left(t\right)+bg\left(t\right)\right]=aL\left\{f\left(t\right)\right\}+bL\left\{f\left(t\right)\right\}$
where a, b are constants. $L\left\{x\left(t\right)\right\}=2L\left\{{t}^{2}\right\}+L\left\{{\int }_{0}^{t}sin\left[2\left(t-\tau \right)\right]x\left(\tau \right)d\tau \right\}$
Step3
By using convolution theorem
$L\left\{{\int }_{0}^{t}\mathrm{sin}\left[2\left(t-\tau \right)\right]x\left(\tau \right)d\tau \right\}=L\left\{\mathrm{sin}\left(2t\right)\right\}L\left\{x\left(t\right)\right\}$
Therefore,
$L\left\{x\left(t\right)\right\}=2L\left\{{t}^{2}\right\}+L\left\{\mathrm{sin}\left(2t\right)\right\}L\left\{x\left(t\right)\right\}$
Use the formula:
$L\left\{{t}^{n}\right\}=\frac{n!}{{s}^{n+1}},L\left\{\mathrm{sin}at\right\}=\frac{a}{{s}^{2}+{a}^{2}}$
It gives
$x\left(s\right)=2\left(\frac{2!}{{s}^{2+1}}\right)+\left(\frac{2}{{s}^{2}+{2}^{2}}\right)x\left(s\right)$
$x\left(s\right)=\frac{4}{{s}^{3}}+\frac{2}{{s}^{2}+4}x\left(s\right)$
Solve above equation for x(s)
$x\left(s\right)-\frac{2}{{s}^{2}+4}x\left(s\right)=\frac{4}{{s}^{3}}$
$\left(1-\frac{2}{{s}^{2}+4}\right)x\left(s\right)=\frac{4}{{s}^{3}}$
$\left(\frac{{s}^{2}+2}{{s}^{2}+4}\right)x\left(s\right)=\frac{4}{{s}^{3}}$
Therefore,
$x\left(s\right)=\frac{4\left({s}^{2}+4\right)}{{s}^{3}\left({s}^{2}+2\right)}$
Step 4
By using partial fractions:
$\frac{4\left({s}^{2}+4\right)}{{s}^{3}\left({s}^{2}+2\right)}=\frac{-2}{s}+\frac{8}{{s}^{3}}+\frac{2s}{{s}^{2}+2}$
Taking inverse Laplace transform on both the sides.
${L}^{-1}\left\{\frac{4\left({s}^{2}+4\right)}{{s}^{3}\left({s}^{2}+2\right)}\right\}={L}^{-1}\left\{\frac{-2}{s}+\frac{8}{{s}^{3}}+\frac{2s}{{s}^{2}+2}\right\}$
Using linearity property of Inverse Laplace transform,
${L}^{-1}\left\{x\left(s\right)\right\}=-2{L}^{-1}\left\{\frac{1}{s}\right\}+8{L}^{-1}\left\{\frac{1}{{s}^{3}}\right\}+2{L}^{-1}\left\{\frac{s}{{s}^{2}+2}\right\}$
Step 5
Use formulae for Inverse Laplace transform,