## 7 Examples  
To clarify the application of the proposed technique and to evaluate its performance, we consider from the literature five example integral equations with known exact solutions and construct approximate explicit solutions in several ways. We emphasize that the solutions obtained with the proposed procedure have an explicit form. However, to avoid listing large expressions and to be able to compare these solutions, except in some cases, we convert all coefficients to floating point numbers with six decimal places without rounding. For the error estimation between the exact solution \(u\) and the approximate solution \(\tilde{u}_{n}\), we use the \(\infty\) norm, i.e.,  
\[\epsilon_{n}=\|u-\tilde{u}_{n}\|_{\infty}=\underset{a\leq x\leq b}{max}|u(x)- \tilde{u}_{n}(x)|.\]  
**Example 1**. : _Let the integral equation [23]_  
\[u(x)-\int_{0}^{\pi}\sin(x-t)u(t)dt=1,\quad x\in[0,\pi]. \tag{42}\]  
_The kernel \(K(x,t)=\sin(x-t)\) and the input function \(f(x)=1\) are continuous on \([0,\pi]\times[0,\pi]\) and \([0,\pi]\), respectively, and we seek the unique solution \(u(x)\in C[0,\pi]\). We solve this equation exactly as well as approximately._  
_DCM: Exact solution:_  
_Since_  
\[K(x,t)=\sin(x-t)=\sin x\cos t-\cos x\sin t,\]  
_i.e., \(K(x,t)\) is separable, the integral equation in (42) is written as_  
\[u(x)-\sin x\int_{0}^{\pi}\cos tu(t)dt+\cos x\int_{0}^{\pi}\sin tu(t)dt=1. \tag{43}\]  
_Following the procedure in Section 3, we set up the vectors_  
\[G=\big{(} \begin{array}{cc}\sin x&-\cos x\end{array} \big{)},\quad\Psi(u)= \left( \begin{array}{c}\int_{0}^{\pi}\cos tu(t)dt\\ \int_{0}^{\pi}\sin tu(t)dt\end{array} \right),\]  
_and write (43) as_  
\[Tu=u-G\Psi(u)=1.\]  
_Condition (5) is fulfilled, specifically \(\det\mathbf{V}=\frac{\pi^{2}}{4}+1\
eq 0\), and thus from (6) it follows that_  
\[u(x)=1-\frac{4\pi}{\pi^{2}+4}\sin x-\frac{8}{\pi^{2}+4}\cos x,\]  
_which is the exact solution of (42)._  
_DKM: Taylor series:_  
_Let us now approximate the kernel \(K(x,t)\) by a truncated Taylor series in \(t\) about the point \(0\) where all terms through \(t^{n}\) are included, i.e.,_  
\[K_{n}(x,t)=\left(\sum_{j=1}^{(n+2)/2}\frac{(-1)^{j-1}t^{2(j-1)}}{(2(j-1))!} \right)\sin x-\left(\sum_{j=1}^{(n+1)/2}\frac{(-1)^{j-1}t^{2j-1}}{(2j-1)!} \right)\cos x.\]  
_As an illustration, let \(n=2\). Then, after substituting \(K_{2}(x,t)\), Equation (42) degenerates to_  
\[\tilde{u}_{2}(x)-\sin x\int_{0}^{\pi}\!\left(1-\frac{t^{2}}{2}\right)\tilde{u }_{2}(t)dt+\cos x\int_{0}^{\pi}t\tilde{u}_{2}(t)dt=1. \tag{44}\]  
_Following the steps in Section 4.1, we specify the vectors_  
\[G=\big{(} \begin{array}{cc}\sin x&-\cos x\end{array} \big{)},\quad\Psi(\tilde {u}_{2})=\left( \begin{array}{c}\int_{0}^{\pi}\!\left(1-\frac{t^{2}}{2} \right)\tilde{u}_{2}(t)dt\\ \int_{0}^{\pi}\tilde{u}_{2}(t)dt\end{array} \right),\]or alternatively_  
\[G=\big{(} \begin{array}{cc}\sin x&-\cos x&-\frac{sinx}{2}\end{array} \big{)}, \quad\Psi(\tilde{u}_{2})=\left( \begin{array}{c}\int_{0}^{\pi}\tilde{u}_{2}(t)dt \\ \int_{0}^{\pi}t\tilde{u}_{2}(t)dt\\ \int_{0}^{\pi}t^{2}\tilde{u}_{2}(t)dt\end{array} \right),\]  
_and put (42) in the form_  
\[T_{s}\tilde{u}_{2}=\tilde{u}_{2}-G\Psi(\tilde{u}_{2})=1.\]  
_Solving by (5) and (6), we acquire_  
\[\tilde{u}_{2}(x)=1-\frac{4\pi^{3}+12\pi}{6\pi^{2}+36}\sin x-\frac{\pi^{4}-6 \pi^{2}}{6\pi^{2}+36}\cos x.\]  
_In a similar manner, other approximate solutions of the same analytic form for higher values of \(n\) are derived. We tabulate some of these solutions in Table 1 and compare them against the exact answer.