convert differential equation to Integral equation $$ y''(x) + y(x) = x$$
with b.v conditions $$ y(0) = 1, y'(1) = 0 $$
Integrating $$ y'(x) - y'(0) + \int \limits _0 ^x y(x)dx = \frac {x^2} 2$$
$ let y'(0) = c_1 $
$$ y'(x) - c_1 + \int \limits _0 ^x y(x)dx = \frac {x^2} 2$$
$$ y'(x) = c_1 - \int \limits _0 ^x y(x)dx + \frac {x^2} 2$$
$$ => c_1 = -\frac {1} 2 + y(1) $$
$$ y'(x) = -\frac {1} 2 + y(1) + \frac {x^2} 2- \int \limits _0 ^x y(x)dx $$
$$ y'(x) = -\frac {1} 2 + c_2 + \frac {x^2} 2- \int \limits _0 ^x y(x)dx $$
again Integrating
$$ y(x) - y(0) = -\frac {x} 2 + c_2x + \frac {x^3} 6- \int \limits _0 ^x \int \limits _0 ^x y(t)dtdx $$
$$ y(x) = \frac {x^3} 6-\frac {x} 2 +1+ c_2x - \int \limits _0 ^x (x-t) y(t)dt $$
further if I put x=0 then $c_2$ will vanish ? and then how could I find the Fredholm I.E from it.

First integral equation must be
$$
y'(x)-y'(0)+\int_{0}^{x}y(s)ds=\frac{x^2}{2}.
$$
Finally you will arrive at
$$
y(x)=y(0)-\frac{1}{2}x+Ax-\int_{0}^{x}\int_{0}^{s}y(t)dtds+\frac{x^{3}}{6},\quad A=\int_{0}^{1}y(s)ds
$$
which satisfies your boundary values.
