What is the most number of zeroes that a solution to y\'\' - [(x^2)+1]y = 0 can

Question

What is the most number of zeroes that a solution to y'' - [(x^2)+1]y = 0 can have?
I'm a little confused because the zeroes of -[(x^2)+1] are not real and therefore there are times when -[(x^2)+1] > 0 and the theorem I have does not apply. Please help!
Instinctively I want to say that there would be at most one zero per solution for a total of two zeroes for the differential equation due to uniqueness of solutions. Also, since the differential equation is second order linear, I instinctively want to think that there are two possible basis vectors and therefore two ways to be zero. Any advice?

Explanation / Answer

I have figured out the answer. Since the equation is of the form y'' + q(x)y = 0 and since q(x) = -[(x^2)+1] < 0 for all x, any nonzero solution to the differential equation has at most one zero.

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