Directions: Find an interval centered about $x = 0$ for which the given initial-value problem has a unique solution.
$$(x - 2)y'' + 3y = x$$
Initial values: $y(0) = 0,\,\,y'(0) = 1 $.
My answer was $(-\infty, 1) \cup (3, \infty)$.
The books answer was $(-\infty < x < 2)$.
I determined my answer because I thought as long as
$a_{2}(x)$(the coefficient of the highest order derivative) $= 0$ is not true.
Maybe it's because I'm not really sure what "centered around $x = 0$" means. But can someone explain to me why the books answer is correct and mine is incorrect.