= Solution
The original PDF shows $x^2$ in the nonzero source branch. Inserting that source into $G_0$ gives, for $x<1/2$,
$$
y_p(x)=\frac x2\int_{1/2}^1(z^2-1)\,dz=-\frac{5x}{48}.
$$
For $x\geq1/2$, split the integral at $z=x$:
$$
\begin{aligned}
y_p(x)&=\frac{x^3-x}{2}\int_{1/2}^x1\,dz+\frac x2\int_x^1(z^2-1)\,dz\\
&=\frac{x^4}{3}-\frac{x^3}{4}-\frac{x}{12}.
\end{aligned}
$$
Because of the homogeneous freedom established in part (i), the full answer to the stated endpoint problem is
$$
\boxed{y(x)=\begin{cases}-5x/48,&0\leq x<1/2,\\x^4/3-x^3/4-x/12,&1/2\leq x\leq1,\end{cases}+C(x^3-x),\qquad C\in\mathbb R.}
$$
At the interface, both branches have value $-5/96$ and <derivative> $-5/48$, so no extra delta or delta <derivative> occurs there. Direct substitution gives zero source on the left and $x^2$ on the right; the endpoint values are zero for every $C$. If the extra condition $y/x\to0$ is imposed, it selects $C=-5/48$, but that is additional information absent from the printed question.
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