Let $u"+p_1(t)u=0$ and $v"+p_2(t)v=0$ with $p_2(t)>p_1(t)$ in $(a,b)$. Suppose that $u(a)=v(a)=0$ and $u'(a)=v'(a)=x>0$. Show that there exists $\epsilon>0$ such that $v(t)>u(t)$ in $(a,a+\epsilon)$.
My attempt:
First with both the equations I write $$u"-v"+(p_1-p_2)(t)(u-v)=0$$ and $$(u-v)(a)=0 , (u-v)'(a)=0$$ So, from an application of sturm theorem that if suppose $p(t)<0$ then the solution of equation $y"+p(t)y=0$ has no more than one zero. I am trying to use this result here. But how to show $(u-v)(t)$ has exactly one zero in $(a,b)$. Please suggest a solution to this!!
Since $u(a)=v(a)=0,u'(a)=v'(a)=x>0$, there is $\epsilon>0$ such that $v(t)>0$ for $t\in(a,a+\epsilon)$. Note $$ \bigg(\frac uv \bigg)'(t)=\frac{u'(t)v(t)-u(t)v(t)'}{v^2(t)}. \tag1$$ Multiplying the 1st equation by $v$ and by integration by parts, one has $$ u'(t)v(t)=\int_a^tu'(s)v'(s)ds-\int_a^tp_1(s)u(s)v(s)ds. \tag2 $$ Similarly, from the 2nd equation, one has $$ u(t)v'(t)=\int_a^tu'(s)v'(s)ds-\int_a^tp_2(s)u(s)v(s)ds. \tag2 $$ From (1), (2) and (3), one has $$ \bigg(\frac uv \bigg)'(t)=\frac{u'(t)v(t)-u(t)v(t)'}{v^2(t)}=-\frac1{v^2(t)}\int_a^t(p_1(s)-p_2(s))u(s)v(s)ds<0 $$ which means that $\bigg(\frac uv \bigg)(t)$ is decreasing in $(a,a+\epsilon)$. So for $s,t\in(a,a+\epsilon)$ ($s<t$), one has $$ \bigg(\frac uv \bigg)(s)>\bigg(\frac uv \bigg)(t). $$ Letting $s\to a$ gives $$ \bigg(\frac uv \bigg)(t)<1 \text{ or }u(t)<v(t). $$