Solution
= Solution
Again $C=\sin(2\pi/3)=\sqrt3/2$, but the principal <inverse sine> would give $\pi/3$ at $x=1$. The required branch is $v=\pi-\arcsin(Cx)$, corresponding to $k=1$. Therefore
$$
\boxed{y(x)=x\left[\pi-\arcsin\left(\frac{\sqrt3x}{2}\right)\right],\qquad0<x<\frac2{\sqrt3}.}
$$
At $x=1$ the bracket is $2\pi/3$, as required. Both the <initial condition> and the original <differential equation> are thus satisfied on the stated branch.