Finding the interval for a initial value problem

Question

I have to find the interval $I$ such that $y:I\rightarrow\mathbb{R}$ is the maximal solution for the initial value problem $$y'+\frac{3}{x}y=x, y(1)=1.$$

I found with the method of variation of constants that $y(x)=\frac{x^2}{5}+\frac{4}{5x^3}$. So for me it seemed obvious that $I=\mathbb{R}\setminus \{0\}$. But when I checked the answer I saw that $I=]0,\infty[$.

Why is this the case?

Answer 1

You have to find an interval, and the set $\mathbb{R}\backslash \{0\}$ is a union of two intervals. Since you are given an initial condition to the right of $0$, the corresponding interval would be $(0, \infty)$ as you have indicated from the answer key.