= Solution
Using the causal <Green's function>,
$$
y_p(x)=\int_0^xG(x,\xi)\xi\,d\xi=x^r\int_0^x\xi^{2-r}\log(x/\xi)\,d\xi.
$$
Put $\xi=xs$. Since $\alpha>-5$ is equivalent to $r<3$, the <improper integral> converges and gives
$$
y_p(x)=x^3\int_0^1s^{2-r}(-\log s)\,ds=\frac{x^3}{(3-r)^2}=\boxed{\frac{4x^3}{(\alpha+5)^2}.}
$$
The elementary identity $\int_0^1s^{p-1}(-\log s)ds=1/p^2$ follows by differentiating $\int_0^1s^{p-1}ds=1/p$ for $p>0$. Direct substitution checks $L(x^3)=(\alpha+5)^2x/4$, and the displayed solution and its first <derivative> vanish at zero.
To give all solutions allowed by the printed <boundary conditions>, retain the singular-endpoint qualification from part (b):
$$
\boxed{y(x)=\begin{cases}\dfrac{4x^3}{(\alpha+5)^2}+x^{(1-\alpha)/2}(C_1+C_2\log x),&-5<\alpha<-1,\\\dfrac{4x^3}{(\alpha+5)^2},&\alpha\geq-1.\end{cases}}
$$
Thus the causal integral gives a particular member of a two-parameter family in the first range and the unique solution in the second.
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