Suppose that we have some smooth $x(t), y(t)$ satisfying $$ x''+Px=0, \; \; y''+Qy=0 $$ with $Q(t)\geq P(t)$ for all $t$ and $x(0)=y(0)=0$, $x'(0)=y'(0)>0$. I would like to show that $x(t)\geq y(t)$ for all $t\geq 0$. I know that this is similar to Sturms comparison theorem which compares the zeros of the functions $x$ and $y$ but how would you go about proving this directly?
I have tried multiplying the first equation by $y$ and the second by $x$ and subtracting to get $$ (x'y-y'x)'=x''y-y''x=(Q-P)xy. $$ Then I can integrate this expression (say from $0$ to $s$) and use the fact that $x(0)=y(0)=0$ to see that $$ x'(s)y(s)-y'(s)x(s)=\int_0^s (Q-P) xy \; dt. $$ I know that $x,y$ are positive for short time after $0$ (since $x'(0)=y'(0)>0$) so I was thinking of using this fact and choose $s$ to be the first 0 of either $x$ or $y$ so that the RHS is positive.
I am not sure if this is the correct approach or if this works at all but this is the only way I see to use the fact that $Q\geq P$. I would love a hint!
My hint is $Q(t)=4\ge1=P(t)$, $x(t)=\sin t$, $y(t)=\frac12\sin2t$, $x(0)=y(0)=1$, $x^{\prime}(0)=y^{\prime}(0)=1>0$ but $y(3\pi/2)=0>x(3\pi/2)=-1$. When $x(t)$ and $y(t)$ are oscillatory it's a bit of a toss-up as to who's on top.