This is just to record that the condition $f\in C^2$ in the accepted answer can be weakened to Lipschitz continuity of $f$. f$, and that$r_{\pm}$do not have to be consecutive zeros. With$y=x'$, let us write the equation as $$y'=f(y).$$ First of all,$y\not\equiv0$because$f(0)\neq0$. On the other hand, if$f(\eta)=0$for some$\eta\neq0$then$y$cannot take the value$\eta$because if$y(b)=\eta$for some$b\in[0,1]$then by uniqueness we would have$y\equiv\eta$. Now, if$y(t)>0$for all$t\in(0,1)$then by definition$x$would be strictly increasing on$(0,1)$, and likewise$y<0$on$(0,1)$implies that$x$would be strictly decreasing on$(0,1)$, meaning that it would not be possible to satisfy the boundary conditions$x(0)=x(1)=0$. Hence$y$must change sign on$(0,1)$, i.e., there is$a\in(0,1)$such that$y(a)=0$. The claim follows from the fact that$y$is a continuous function that takes the value$0$, but cannot take the values$r_{\pm}$. 1 This is just to record that the condition$f\in C^2$in the accepted answer can be weakened to Lipschitz continuity of$f$. With$y=x'$, let us write the equation as $$y'=f(y).$$ First of all,$y\not\equiv0$because$f(0)\neq0$. On the other hand, if$f(\eta)=0$for some$\eta\neq0$then$y$cannot take the value$\eta$because if$y(b)=\eta$for some$b\in[0,1]$then by uniqueness we would have$y\equiv\eta$. Now, if$y(t)>0$for all$t\in(0,1)$then by definition$x$would be strictly increasing on$(0,1)$, and likewise$y<0$on$(0,1)$implies that$x$would be strictly decreasing on$(0,1)$, meaning that it would not be possible to satisfy the boundary conditions$x(0)=x(1)=0$. Hence$y$must change sign on$(0,1)$, i.e., there is$a\in(0,1)$such that$y(a)=0$. The claim follows from the fact that$y$is a continuous function that takes the value$0$, but cannot take the values$r_{\pm}\$.