Sturm Comparison Theorem 
For part (c) my book says "But clearly \begin{align*} &\phi_2(a) \geq 0, \phi_2(b) \geq 0 \\&\phi_1'(a) \geq 0, \phi_1'(b) \leq 0."\end{align*} I don't see how they are getting this. We aren't given information on the sign of $\phi_1,\phi_2$ if $x \not \in (a,b)$ are we?
 A: Note that $\phi_1(x) > 0$ for all $x \in (a,b)$ and that $\phi_1$ is continuous (since it's twice differentiable). So 
$$\lim_{x \, \searrow \,a} \phi_1(x) \geq 0$$
as it is a limit of strictly positive numbers. The same reasoning applies to $\phi_2$ and at $b$. So we now know that $\phi_1(a) \geq 0, \phi_2(b) \geq 0$.
Now suppose that $\phi_1(a) = \phi_1(b) = 0$. This is to set up a proof by contradiction. Then from part $(b)$, you get the equation
$$\phi_1'(b)\phi_2(b) > \phi_1'(a)\phi_2(a).$$
Since $\phi_2(b) \geq 0$ and $\phi_2(a) \geq 0$, we must have that $\phi_1'(b) > \phi_1'(a).$
On the other hand, since $\phi_1(a) = 0$ and $\phi_1(x) > 0$ for $a < x < b$, we must have that $\phi_1'(a) \geq 0$ (since it's increasing there). Similarly, $\phi_1(b) = 0$ and $\phi_1(x) > 0$ for $a < x < b$, so $\phi_1'(b) \leq 0$ (since it's decreasing there).
This shows that $\phi_1'(b) \leq \phi_1'(a)$, which directly contradicts the inequality above.
So it cannot happen that $\phi_1(a) = \phi_2(b) = 0$.
One can get part $(d)$ immediately from this by replacing $\phi_1$ (respectively $\phi_2$) by $-\phi_1$ (respectively $-\phi_2$) when appropriate. $\diamondsuit$
