= Solution
The substitution $y=xv$ gives $y'=v+xv'$. Cancelling $x^3v$ on both sides of the <differential equation>, for $x>0$, gives
$$
\boxed{xv'=v^3}.
$$
The <initial condition> is $v(1)=1$. This selects a positive solution locally, so <separation of variables> gives
$$
\int v^{-3}\,dv=\int x^{-1}\,dx,
\qquad -\frac1{2v^2}=\log x+C.
$$
Using $v(1)=1$ gives $C=-1/2$, and continuity selects the positive square root:
$$
v=\frac1{\sqrt{1-2\log x}},\qquad
\boxed{y=\frac{x}{\sqrt{1-2\log x}}=\sqrt{\frac{x^2}{1-\log(x^2)}},\quad m=2}.
$$
The real solution containing $x=1$ has maximal positive domain \b[$0<x<\sqrt e$]. Its denominator tends to zero and $y\to+\infty$ as $x\uparrow\sqrt e$, a <finite-time blow-up of an ordinary differential equation> in the independent variable $x$. Thus the source's wording “for $x>0$” needs this domain restriction; the prescribed initial value admits no real solution on the whole positive half-line. The division by $v^3$ loses the separate solution $v=0$, but that solution does not meet the <initial condition>.
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