= Solution
Multiplying the <ordinary differential equation> by two suggests using the <dependent variable> $Y=y^2$, whose <derivative> is $Y'=2yy'$. On the interval through $x=1$, this gives the <linear ordinary differential equation>
$$
xY'=Y+x^2,\qquad \left(\frac Yx\right)'=1.
$$
After <integration>, $Y=x^2+Cx$. The <initial condition> $Y(1)=4$ gives $C=3$. As $y(1)>0$, <continuity> selects the positive branch:
$$
\boxed{y(x)=\sqrt{x^2+3x}\quad(x>0),\qquad y(2)=\sqrt{10}.}
$$
The interval $x>0$ is the maximal real interval through the given <initial condition>: at $x=0$ the formula approaches $0$, but its <derivative> is unbounded, so it does not extend through zero as a differentiable <solution of a differential equation>.
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