determine the longest interval in which the given initial value problem is certain to have a unique twice differentiable solution without finding the solution.

ty" + 3y = t, y(1) = 1, y'(1)=2

i set this up and got

y" + (3/t)y = 1

the interval in the book says its 0< t< infinity, but im not understanding that.

When I look at q(t) = (3/t) it looks like it would have a continuous solution everywhere except zero. why are they truncating the negative?