Question : What is solution of the integral equation
\[y\left(x\right)=x^3+\int_{0}^{x}\sin{\left(x-t\right)}y\left(t\right)dt\ ,\ x\in[0,\ \pi]\]
Solution :
Given integral equation
\[y\left(x\right)=x^3+\int_{0}^{x}\sin{\left(x-t\right)}y\left(t\right)dt\ ,\ x\in[0,\ \pi]\]
Can be written as
\[y\left(x\right)=x^3+k\left(x\right)\ast y(x)\]
Where k(x)*v(x) is convolution of u and v functions
Now applying Laplace transform both sides
\[Y\left(s\right)=\frac{6}{s^4}+K\left(s\right)\ Y(s)\]Where K(s) =Laplace transform of sin(x) = 1/(s^2+1)
\[Y\left(s\right)=\frac{6}{s^4}+\frac{Y\left(s\right)}{s^2+1}\] \[=>\left(1-\frac{1}{s^2+1})\right)Y\left(s\right)=\frac{6}{s^7} \]
\[=>\left(\frac{s^2}{s^2+1}\right)Y\left(s\right)=\frac{6}{s^4}\]
\[=>\ Y\left(s\right)=\frac{6\left(s^2+1\right)}{s^6}=6\left(\frac{1}{s^4}+\frac{1}{s^6}\right)\]
\[=>Y\left(s\right)=6\left(\frac{x^3}{6}+\frac{x^5}{120}\right)=x^3+\frac{x^5}{20}\]Applying inverse Laplace transform both sides
\[=>y\left(x\right)=x^3+\frac{x^5}{20}\]is the required solution of the integral equation.
and the value of y(1) is obtained by substituting
\[y\left(1\right)=1+\frac{1}{20}=\frac{21}{20}\]