## 4 Examples  
To show the application and the effectiveness of the method described in the previous Section, we solve two illustrative example boundary value problems. First, an ordinary integro-differential equation with nonlocal boundary conditions is considered. Next, an integro-partial differential equation with integral conditions is factorized and solved in closed form. In the Appendix A, we present the implementation of the procedure in the open source software Maxima. **Example 1**. : _Consider the ordinary integro-differential equation_  
\[u^{\prime\prime\prime}(x)-8x^{2}\int_{0}^{1}tu^{\prime}(t)dt-(3x+1)\int_{0}^{ 1}t^{2}u^{\prime\prime\prime}(t)dt=f(x),\quad 0<x<1, \tag{20}\]  
_subject to nonlocal boundary conditions_  
\[u(0)=2\int_{0}^{1}u(t)dt,\quad u^{\prime}(0)=-u^{\prime}(1),\quad u^{\prime \prime}(0)=-u^{\prime\prime}(1), \tag{21}\]  
_for any \(f(x)\in C[0,1]\)._  
1. _According to procedure in Listing 1, we take_ \(X=C[0,1]\) _and define the operator_ \(B:C[0,1]\to C[0,1]\) _by_ \[Bu = u^{\prime\prime\prime}(x)-8x^{2}\int_{0}^{1}tu^{\prime}(t)dt-(3x+ 1)\int_{0}^{1}t^{2}u^{\prime\prime\prime}(t)dt,\] \[D(B) = \bigg{\{}u\in C^{3}[0,1]:u(0)=2\int_{0}^{1}u(t)dt,\;u^{\prime}(0 )=-u^{\prime}(1),\;u^{\prime\prime}(0)=-u^{\prime\prime}(1)\bigg{\}}.\]_Further, we define the operators_ \(A,A_{1}:C[0,1]\to C[0,1]\) _of order_ \(n=3\) _and_ \(n_{1}=1\)_, respectively, as_ \[Au = u^{\prime\prime\prime}(x),\quad D(A)=D(B),\] \[A_{1}u = u^{\prime}(x),\quad D(A_{1})=\bigg{\{}u\in C^{1}[0,1]:u(0)=2\int_{ 0}^{1}u(t)dt\bigg{\}}.\]
3. _The operator_ \(A_{2}:C[0,1]\to C[0,1]\) _is determined by requiring_ \[Au = u^{\prime\prime\prime}(x)=A_{2}A_{1}u=A_{2}(u^{\prime}(x))=A_{2 }v=v^{\prime\prime}(x),\quad v=u^{\prime}(x),\] \[D(A_{2}A_{1}) = \big{\{}u\in D(A_{1}):A_{1}u=u^{\prime}(x)=v\in D(A_{2})\big{\}}\] (22) \[= \bigg{\{}u\in C^{1}[0,1]:u(0)=2\int_{0}^{1}u(t)dt,\;u^{\prime}(x) =v\in D(A_{2})\bigg{\}}\] \[= D(B),\] _which implies that_ \[u(0)=2\int_{0}^{1}u(t)dt,\quad u^{\prime}(0)=-u^{\prime}(1),\quad u^{\prime \prime}(0)=-u^{\prime\prime}(1), \tag{21}\]0
4. _The operators_ \(A_{1}\) _and_ \(A_{2}\) _are known to be correct and their inverses for any_ \(f(x)\in C[0,1]\) _are given by_ \[u(0)=2\int_{0}^{1}u(t)dt,\quad u^{\prime}(0)=-u^{\prime}(1),\quad u^{\prime \prime}(0)=-u^{\prime\prime}(1), \tag{21}\]1 \[u(0)=2\int_{0}^{1}u(t)dt,\quad u^{\prime}(0)=-u^{\prime}(1),\quad u^{\prime \prime}(0)=-u^{\prime\prime}(1), \tag{21}\]2
5. _For_ \(m=1\)_, we set up the vectors_ \[u(0)=2\int_{0}^{1}u(t)dt,\quad u^{\prime}(0)=-u^{\prime}(1),\quad u^{\prime \prime}(0)=-u^{\prime\prime}(1), \tag{21}\]3 \[u(0)=2\int_{0}^{1}u(t)dt,\quad u^{\prime}(0)=-u^{\prime}(1),\quad u^{\prime \prime}(0)=-u^{\prime\prime}(1), \tag{21}\]4
6. _We compute_ \[u(0)=2\int_{0}^{1}u(t)dt,\quad u^{\prime}(0)=-u^{\prime}(1),\quad u^{\prime \prime}(0)=-u^{\prime\prime}(1), \tag{21}\]5 (23)
7. _By utilizing (_13_), we find_ \[u(0)=2\int_{0}^{1}u(t)dt,\quad u^{\prime}(0)=-u^{\prime}(1),\quad u^{\prime \prime}(0)=-u^{\prime\prime}(1), \tag{21}\]6 (24)
8. _Finally, we compute_ \[u(0)=2\int_{0}^{1}u(t)dt,\quad u^{\prime}(0)=-u^{\prime}(1),\quad u^{\prime \prime}(0)=-u^{\prime\prime}(1), \tag{21}\]7 (25)
9. _From (_22_)-(_25_) it is implied that all requirements of Theorem_ 2 _are satisfied and therefore the given boundary value problem (_20_), (_21_) can be factorized and its unique solution may be obtained by substituting into (_17_). As an illustration, for_ \(f(x)=2x^{2}-6x+4\) _the unique solution turns out to be_ \[u(0)=2\int_{0}^{1}u(t)dt,\quad u^{\prime}(0)=-u^{\prime}(1),\quad u^{\prime \prime}(0)=-u^{\prime\prime}(1), \tag{21}\]8  
**Example 2**.