# IE13

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

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

where

$k(x,u)={\begin{cases}u,\,0