= Solution
Choose $y_a=x$ and $y_b=x^3-x$, satisfying the left and right conditions respectively. Their <Wronskian> is
$$
W(x)=x(3x^2-1)-(x^3-x)=2x^3.
$$
Part (a) therefore supplies the valid kernel
$$
\boxed{G_0(x,z)=\begin{cases}\dfrac{x(z^3-z)}{2z^3},&0\leq x\leq z,\\[4pt]\dfrac{x^3-x}{2z^2},&z\leq x\leq1,\end{cases}\qquad0<z<1.}
$$
It is continuous at the source, its first <derivative> has jump one, and it vanishes at both endpoints. However, it is not unique: $x^3-x$ is a nonzero homogeneous solution satisfying both stated endpoint conditions. Thus $G_0+(x^3-x)h(z)$ has the same defining properties for any suitable $h$.
This is a <nonunique Green kernel at a singular endpoint with two vanishing modes>. Choosing $y_a=x^3$ instead also gives a valid kernel, corresponding to the stronger selection $y(x)/x\to0$ at the left endpoint. The original problem states no such selection, so the family must be acknowledged.
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