# Determining properties of solution of a second order ODE

If the functions $y_1$ and $y_2$ are linearly independent solutions of $$y'' + p(t)y' + q(t)y = 0$$, show that between consecutive zeros of $y_1$ there is one and only one zero of $y_2$. Note that this

result is illustrated by the solutions $y_1(t) = cos t$ and $y_2(t) = sin t$ of the equation $y'' + y = 0$.

What i tried

I dont quite get what the question means when it says between consecutive zeros of $y_1$

there is one and only one zero of $y_2$. What i do know is that since $y_1$ and $y_2$ are

linearly independent, then the linear combination of $y_1$ and $y_2$ can be written as

$$c_1y_1+c_2y_2=0$$

Where $c_1=c_2=0$

Could anyone explain. Thanks

• Consecutive zeros mean between the points of $\cos t =0$ namely $$x= \left(n-\frac{1}{2}\right)\pi$$ plus what is the shift between $\cos t$ and $\sin t$? – Chinny84 Feb 20 '15 at 16:42
• cost=sin(t+$\pi/2$) – ys wong Feb 20 '15 at 16:53
• i still dont get it could anyone explain. thanks – ys wong Feb 20 '15 at 17:25
• i posted an explanation. check it out. – abel Feb 20 '15 at 20:04

i think you can show this interlacing property by considering the wronskian $w.$
suppose we have two linearly independent solutions $y_1$ and $y_2.$ suppose too, that there are two points $a < b$ such that $y_1(a) = 0, y_1(b) = 0, y_1(x) > 0 \text{ for } a < x < b \text{ and wlog } y_2(a) > 0.$
let the wronskian $w$ be defined by $$w(x) = y_1'(x)\,y_2(x) - y_2'(x)y_1(x).$$ we will show that $y_2$ has a zero in $(a, b)$ by showing that $y_2(a)y_2(b) < 0.$ note that by uniqueness theorem $y'_1(a) \neq 0$ and by the positivity of $y_1,$ we have in fact $y'_1(a) > 0$ and in the same way $y'_1(b) < 0.$ that is $$y'_1(a) > 0 \text{ and } y'_1(b) < 0. \tag 1$$
by abel's theorem, $$w = w(a)e^{-\int_a^x p(t) \, dt}.$$ that is $w(x)$ has the same sign as $w(a)$ for $a < x < b.$ now $$w(a) = y'_1(a)y_2(a)-y'_2(a)y_1(a)=y'_1(a)y_2(a) > 0. \tag2$$
and $$w(b) = y'_1(b)y_2(b)-y'_2(b)y_1(b)=y'_1(b)y_2(b) > 0.\tag3$$
$(1), (2)$ and $(3)$ proves the claim.