Solution
= Solution
The <initial condition> gives $C=\sin(\pi/3)=\sqrt3/2$. Since $\pi/3$ belongs to the principal <inverse sine> range, choose $k=0$ in part (i). The particular solution is
$$
\boxed{y(x)=x\arcsin\left(\frac{\sqrt3x}{2}\right),\qquad0<x<\frac2{\sqrt3}.}
$$
The upper endpoint is excluded because $y/x\to\pi/2$ there and the original <tangent> ceases to be defined.