maximum Difference between two zeros

Question

What is the maximum difference between the two consecutive zeros of the solutions of $y''+(1+x)y=0$ on $0\leq x<+\infty$? I have applied the Strum's comparison theorem (by comparison with $y''+y=0$), and I obtained this maximum as $2\pi$ but I think it may be $\pi$. Is it necessarily true? (According to the Strum comparison theorem we know that the solution of the above equation have many infinite zeros on $R$).

Answer 1

Equation $$y''+y=0$$ has a general solution $$y=a\cos t + b\sin t$$ Which obtain zero at $$t=\mathrm{atan2}\left(\pm\frac{a}{\sqrt{a^2+b^2}},\mp\frac{b}{\sqrt{a^2+b^2}}\right)=\begin{cases} \arctan\frac{b}{\pm a}&\pm a>0\\ -\pi+\arctan\frac{b}{a}&\pm a<0,\mp b<0\\ \pi+\arctan\frac{b}{a}&\pm a<0,\mp b>0\\ \end{cases}$$ where $\mathrm{atan2}$ is a special version of $\arctan$ that work in all quadrants.

For example $a=1=b$ you have difference between zeros $=\pi$.

---

The situation for $y''+(1+x)=0$ is much more complicate if $y=y(x)$

$$y=c_1 \text{Ai}\left(-\sqrt[3]{-1} (-x-1)\right)+c_2 \text{Bi}\left(-\sqrt[3]{-1} (-x-1)\right)$$ See

http://mathworld.wolfram.com/AiryFunctionZeros.html

http://onlinelibrary.wiley.com/doi/10.1002/zamm.19960760714/abstract