= Solution
Integrate the slope in part (i), using $y(-\alpha)=0$. On the left,
$$
y(x)=\int_{-\alpha}^x(1-s)^{-1/3}\,ds
=\frac32\big[(1+\alpha)^{2/3}-(1-x)^{2/3}\big].
$$
Its value at the interface is $y(0)=\frac32[(1+\alpha)^{2/3}-1]$. On the right, integrate $(1+x)^{-1/2}$ from the interface and enforce continuity. Therefore
$$
\boxed{y(x)=\begin{cases}
\displaystyle\frac32\big[(1+\alpha)^{2/3}-(1-x)^{2/3}\big],&-\alpha\leq x<0,\\[5pt]
\displaystyle\frac32\big[(1+\alpha)^{2/3}-1\big]+2\big[\sqrt{1+x}-1\big],&0\leq x\leq\alpha.
\end{cases}}
$$
Both $y$ and $y'$ are continuous at zero. The ray steepens as it approaches the interface from the left and flattens as it leaves to the right.
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