Proof of Single Zero Between Consecutive Zeros of y1 & y2

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Supose y1 and y2 are a fundamental set of solutions to a second order ODE on the interval
-infinity<t<infinity.
How can I show that there is one and only one zero of y1 between consecutive zeros of y2.

I really don't even know how to egin approaching this problem. Any direction would be greatly appreciated.
 
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only one zero of y1 between consecutive zeros of y2? what do you mean by this statement?
 
I believe that I mean if y1(x)=0 and y1(z)=0 then y2(a) must be equal to zero with x<a<z.

In other words, y1 and y2 must alternate zeros as t varies.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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