= Solution
Trying $y=x^m$ in the <Cauchy-Euler equation> gives
$$
m(m-1)+4m-4=(m-1)(m+4)=0,
$$
so the <homogeneous solution> is
$$
\boxed{y=C_1x+C_2x^{-4}.}
$$
Continuity with zero value at the singular endpoint excludes the second branch. Use the left and right adapted solutions $\ell(x)=x$, $r(x)=x^{-4}-x$, whose <Wronskian> is
$$
W(x)=\ell r'-\ell'r=-5x^{-4}.
$$
The <Dirichlet Green kernel for an Euler operator with exponents one and minus four> is therefore
$$
\boxed{G(x,\xi)=\frac15\begin{cases}x(\xi^5-1),&x\leq\xi,\\\xi^5(x-x^{-4}),&x\geq\xi.\end{cases}}
$$
It vanishes at both endpoints, is continuous at $x=\xi$, and has first-derivative jump
$$
G_x(\xi^+,\xi)-G_x(\xi^-,\xi)=\frac{\ell(\xi)r'(\xi)-\ell'(\xi)r(\xi)}{W(\xi)}=1.
$$
Thus $L_xG=\delta(x-\xi)$ with ordinary source measure $d\xi$. For bounded forcing the <Green function> representation is
$$
\boxed{u(x)=\frac{x-x^{-4}}5\int_0^x\xi^5f(\xi)d\xi+\frac x5\int_x^1(\xi^5-1)f(\xi)d\xi.}
$$
The apparent singular prefactor in the first term is canceled by the integral's decay at zero. The only regular homogeneous solution vanishing at $1$ is zero, so this solution is unique in the stated regular class.
For the piecewise source, split the integrals at $1/2$. If $x<1/2$, the needed integrals are $\int_0^x\xi^6d\xi=x^7/7$ and $\int_x^{1/2}(\xi^6-\xi)d\xi$. If $x>1/2$, only $\int_0^{1/2}\xi^6d\xi=1/896$ remains. Simplification yields
$$
\boxed{u(x)=\begin{cases}\displaystyle\frac{x^3}{14}-\frac{111x}{4480},&0\leq x\leq\frac12,\\\displaystyle\frac{x-x^{-4}}{4480},&\frac12\leq x\leq1.\end{cases}}
$$
At the interface both expressions have value $-31/8960$ and derivative $129/4480$, so no unintended <Dirac delta> source appears. Their second derivatives have the allowed jump associated with the discontinuous forcing.
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