Show $y(x) = \int_{0}^{1}G(x, \xi)g(\xi)d\xi$ We have boundary-value problem$$y'' = g(x)\; y(0)=y(1)=0,$$
Show that:$$y(x) = \int_{0}^{1}G(x, \xi)g(\xi)d\xi$$
with $$G(x,\xi) = \begin{cases} \xi(x-1) & \text{for } 0 \leq \xi \leq x \leq 1 \\ x(\xi-1)) & \text{for } \ \leq x \leq \xi \leq 1\end{cases}$$
Can someone explain to me how to start this problem because, I don't know where to begin with the conditions. 
 A: Start with greens method. That kernal looks like a greens function.
$$
\frac{d^2}{dx^2}G(\zeta,x)  = \delta(\zeta -x)
$$
Multiple by $g(\zeta)$ we find
$$
\frac{d^2}{dx^2}G(\zeta,x) g(\zeta) = \delta(\zeta -x)g(\zeta)
$$
Integrate over $\zeta$ which is bounded between 0 and 1.
Thus
$$
\int_0^1\frac{d^2}{dx^2}G(\zeta,x) g(\zeta)d\zeta= \frac{d^2}{dx^2}\int_0^1G(\zeta,x) g(\zeta)d\zeta = \int_0^1\delta(\zeta -x)g(\zeta)
$$
The later integral is delta function integral so the only non zero component is when $\zeta = x$ thus
$$
\frac{d^2}{dx^2}\int_0^1G(\zeta,x) g(\zeta)d\zeta = g(x)
$$
Compare with your original equation shows 
$$
y(x)  = \int_0^1G(\zeta,x) g(\zeta)d\zeta
$$
A: You can use this way to show. Integrating the equation, you can get
$$ y'(x)=\int_0^xg(t)dt+C_1. $$
Integrating the equation above, you can get
$$ y(x)=\int_0^x\int_0^sg(t)dtds+C_1x+C_2. $$
Using $y(0)=0,y(1)=0$ gives
$$ C_2=0, C_1=-\int_0^1\int_0^sg(t)dtds. $$
Thus
\begin{eqnarray}
y(x)&=&\int_0^x\int_0^sg(t)dtds-x\int_0^1\int_0^sg(t)dtds\\
&=&\int_0^x\int_t^xg(t)dsdt-x\int_0^1\int_t^1g(t)dsdt\\
&=&\int_0^x(x-t)g(t)dt-\int_0^1x(1-t)g(t)dt\\
&=&\int_0^x[(x-t)-x(1-t)]g(t)dt-\int_x^1x(1-t)g(t)dt\\
&=&\int_0^xt(x-1)g(t)dt+\int_x^1x(t-1)g(t)dt\\
&=&\int_0^1G(t,x)g(t)dt.
\end{eqnarray}
