# Using Convolution, Solve the Volterra Integral Equation: y(x)=1-e^{-x}+int_0^x y(x-t)sin x dt

Using Convolution, Solve the Volterra Integral Equation:
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Step 1
Given:

$Y\left(s\right)=\frac{1}{s}-\frac{1}{s+1}+Y\left(s\right)L\left(\mathrm{sin}x\right)$
$Y\left(s\right)=\frac{1}{s}-\frac{1}{s+1}+Y\left(s\right)\frac{1}{1+{s}^{2}}$
$Y\left(s\right)=\left(1-\frac{1}{{s}^{2}+1}\right)=\frac{1}{s\left(s+1\right)}$
$Y\left(s\right)=\frac{{s}^{2}+1}{{s}^{3}\left(s+1\right)}=\frac{\left(s+1{\right)}^{2}-2s}{{s}^{3}\left(s+1\right)}$
$Y\left(s\right)=\frac{s+1}{{s}^{3}}-\frac{2}{{s}^{2}\left(s+1\right)}$
$Y\left(s\right)=\frac{1}{{s}^{2}}+\frac{1}{{s}^{3}}-\frac{2}{{s}^{2}\left(s+1\right)}$
$\text{Step 2}$
$\text{Taking Inverse Laplace Transformation}$
$y\left(x\right)={L}^{-1}\left\{\frac{1}{{s}^{2}}\right\}+{L}^{-1}\left\{\frac{2}{{s}^{2+1}}\right\}-2{L}^{-1}\left\{\frac{1}{{s}^{2}\left(s+1\right)}\right\}$
$=x+\frac{1}{2}{x}^{2}+{\int }_{0}^{x}{\int }_{0}^{x}\left({e}^{-x}dx\right)dx$
$=x+\frac{1}{2}{x}^{2}+{\int }_{0}^{x}\left({e}^{x}-1\right)dx$
$=x+\frac{1}{2}{x}^{2}-{\left[{e}^{-x}+x\right]}_{0}^{x}$
$=x+\frac{1}{2}{x}^{2}-{e}^{-x}-x+1$
$y\left(x\right)=x+\frac{1}{2}{x}^{2}-{e}^{-x}$