Solution
= Solution
On $x\geq1$, an <integrating factor> is $1/x$. Multiplication gives $(y/x)'=1$, so the <general solution> is $y=x^2+Cx$. The initial value $y(1)=1$ forces $C=0$. Therefore
$$
\boxed{y=x^2,\qquad x\geq1}.
$$
= Solution
On $x\geq1$, an <integrating factor> is $1/x$. Multiplication gives $(y/x)'=1$, so the <general solution> is $y=x^2+Cx$. The initial value $y(1)=1$ forces $C=0$. Therefore
$$
\boxed{y=x^2,\qquad x\geq1}.
$$