Solve the BVP
$$y'' = f(x) = x, \qquad y(0) = 0, \qquad y'(1) = 0$$
with the usage of Green's function.
First of all, the solution should be
$$y = \frac{x^3}{6} - \frac{x}{2}.$$
According to this explanation, for calculating $G$, we first need a fundamental system of the associated homogenous equation $y'' = 0$. We take $u_1 = x$ and $u_2 = 1$. According to the boundary values, we now choose $y_1 = x$ and $y_2 = 1$, then we have $y_1(0) = 0$ and $y_2'(1) = 0$. Furthermore, $W(y_1, y_2) = -1$, so it follows
$$G(x,s) = \begin{cases} -s \qquad &0 \leq s \leq x \\ -x \qquad &x \leq s \leq 1. \end{cases}$$
Then I obtain for the solution
$$y = \int_0^1 G(x,s) f(s) \ \mathrm{d}s = -\frac{1}{3},$$
which is obviously not the correct answer. Where is my mistake?
\begin{align} y &= \int_0^1 G(x,s) f(s) \ \mathrm{d}s \\ &= \int_0^x (-s) s \ \mathrm{d}s + \int_x^1 (-x) s \ \mathrm{d}s \\ &= -\frac{x^3}3-\frac{x}2(1-x^2)=\frac{x^3}6-\frac{x}2 \end{align} which seems to be correct. What were the steps that led you to the wrong solution?