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Transform the BVP to an integral equation:{\frac  {d^{2}y}{dx^{2}}}+y=x,y(0)=0,y'(1)=0

First, integrate both sides from x to 1 (so that the boundary condition y'(1)=0 can be applied). This gives
y'(1)-y'(x)+\int _{x}^{1}y(u)\,du=\int _{x}^{1}u\,du={\frac  {u^{2}}{2}}{\Big |}_{x}^{1}={\frac  {1}{2}}-{\frac  {x^{2}}{2}}

Applying the condition y'(1)=0 and multiplying both sides by (-1) results in

y'(x)-\int _{x}^{1}y(u)\,du={\frac  {x^{2}}{2}}-{\frac  {1}{2}}

Next, integrate both sides from 0 to x:

{\big (}y(x)-y(0){\big )}-\int _{0}^{x}\int _{z}^{1}y(u)\,du\,dz=\int _{0}^{x}{\frac  {u^{2}}{2}}-{\frac  {1}{2}}\,du=\left({\frac  {u^{3}}{6}}-{\frac  {u}{2}}\right){\Big |}_{0}^{x}={\frac  {x^{3}}{6}}-{\frac  {x}{2}}

or, applying the condition y(0)=0,

y(x)-\int _{0}^{x}\int _{z}^{1}y(u)\,du\,dz={\frac  {x^{3}}{6}}-{\frac  {x}{2}}

The double integral may be simplified if we invert the order of integration. To do this, consider the region of integration for the double integral:
This same region, when the order of integration is reversed, looks like this:

We see that the region must be split into two pieces, and a separate integral written for each piece:

\int _{0}^{x}\int _{z}^{1}y(u)\,du\,dz=\int _{0}^{x}\int _{0}^{u}dz\,\,y(u)\,du+\int _{x}^{1}\int _{0}^{x}dz\,\,y(u)\,du

or after doing the trivial z-integration,

\int _{0}^{x}\int _{z}^{1}y(u)\,du\,dz=\int _{0}^{x}u\,y(u)\,du+\int _{x}^{1}x\,y(u)\,du

The desired integral equation, then, is

y(x)-\int _{0}^{x}u\,y(u)\,du-\int _{x}^{1}x\,y(u)\,du={\frac  {x^{3}}{6}}-{\frac  {x}{2}}

Such an integral equation is often written in the form

y(x)-\int _{0}^{1}k(x,u)\,y(u)\,du={\frac  {x^{3}}{6}}-{\frac  {x}{2}}