Properties of $y$ if $\frac{d^2y}{dx^2}+xy=0$, $x \in [a,b],$ and $y(a)=y(b)=0.$ Let $y$ be a nontrivial solution of the boundary value problem
$$\frac{d^2y}{dx^2}+xy=0,\ x \in [a,b]$$
$$y(a)=y(b)=0$$ Then:


*

*$b>0$

*$y$ is monotone in $(a,0)$ if $a<0<b$.

*$y'(a)=0$.

*$y$ has infinitely many zeros in $[a,b]$


I have no idea how to solve it and which options is/ are correct. please help me.
 A: This equation with a minus instead of the plus is known as the Airy equation, and it's solution is a linear combination of the the two Airy functions. You can easily show that the solution for:
$$y''(x) -kxy = 0$$
Is therefore:
$$y(x) = c_1\mathrm{Ai}(\sqrt[3]{k}x) + c_2\mathrm{Bi}(\sqrt[3]{k}x)$$
Where in your case, $k=-1$, and we can choose the real root to give:
$$y(x) = c_1\mathrm{Ai}(-x) + c_2\mathrm{Bi}(-x)$$
We also have that:
$$y(a) = c_1\mathrm{Ai}(-a) + c_2\mathrm{Bi}(-a)=0$$
$$y(b) = c_1\mathrm{Ai}(-b) + c_2\mathrm{Bi}(-b)=0$$
If the determinant of the above system is non zero, we get a trivial solution, which we are given is not the case. Thus, we must have:
$$\frac{\mathrm{Bi}(-b)}{\mathrm{Ai}(-b)} = \frac{\mathrm{Bi}(-a)}{\mathrm{Ai}(-a)}$$
Which puts a constraint on our $a,b$. And that our solution is of the form:
$$y(x) = c_1\left(\mathrm{Ai}(-x) - \frac{\mathrm{Ai}(-a)}{\mathrm{Bi}(-a)}\mathrm{Bi}(-x)\right)$$
Now let's look at $\mathrm{A_i}, \mathrm{B_i}$:

Let's start with option number two. If $a<0<b$, that means that $-b<0<-a$. Moreover, $0<\mathrm{Ai}(-a)/\mathrm{Bi}(-a) < 1$ Thus, you can easily see that in the range $(a, 0)$ our function is a positive sum of two monotonically decreasing functions, and is therefore monotonically decreasing itself. Thus, option #2 is true.
As to option #1 (Thanks @zhw!) this is also true. to see this, let's look at the function $\mathrm{Ai}(-x)/ \mathrm{Bi}(-x)$. This function is monotonically increasing for all $x<0$, as can be proved by looking at the derivative:
$$\frac{d}{dx}\left(\frac{\mathrm{Ai}(-x)}{\mathrm{Bi}(-x)}\right)=\frac{\mathrm{Ai}(-x) \mathrm{Bi}'(-x) - \mathrm{Ai}'(-x) \mathrm{Bi}(-x)}{\mathrm{Bi}(-x)^2}=\frac{c}{\mathrm{Bi}(-x)^2}>0$$
So if $b$ were smaller than zero, we cannot satisfy the above condition on the ratios of functions. Thus, option #1 is also true.
As to options 4, if e.g. $a=b$ we cannot have an infinite amount of zeros. So, option #4 is false.
A: $1.$ $b>0$ is true. Suppose to the contrary that $b\le 0.$ Observe that the set $\{x\in [a,b]: y(x)\ne 0\}$ is open in $(a,b),$ hence is the pairwise disjoint union of open intervals. It follows that there exist $a\le a_1<b_1\le b\le 0$ such that $y(a_1)= y(b_1) = 0$ and $y\ne 0$ in $(a_1,b_1).$ Two cases: i) $y>0$ on $(a_1,b_1);$ ii) $y<0$ on $(a_1,b_1).$ In case i), because $x<0,$ we have $y''>0$ on $(a_1,b_1).$ Thus we have a strictly convex function on $[a_1,b_1]$ that vanishes at the endpoints and is positive on $(a_1,b_1).$ That's a clear contradiction. Case ii) similarly leads to a contradiction (or just consider $-y$).
$2.$ This one is true too. The first observation is that $y \ne 0$ on $(a,0).$ Otherwise we end up with an interval $(a_1,b_1)$ as in 1. So we must have $y>0$ on $(a,0)$ or $y<0$ on $(a,0).$  In the first case we have $y'(a) \ge 0.$ But $y>0,x<0$ implies $y''>0$ on $(0,a).$ This implies $y'$ is strictly increasing on $(a,0),$ hence $y'>0$ on $(a,0].$ Thus $y$ is strictly increasing on $(a,0).$ The second case follows similarly (or just consider $-y$).
$3$. and 4. are both false. To see this, let $a<0$ and consider the second order linear ODE on $\mathbb R$ given by $y''+xy=0$ with initial conditions $y(a) = 0,y'(a)=1.$ By the standard existence/uniqueness theorem, this equation has a unique solution on $\mathbb R.$ From now on, $y$ will denote this solution.
As we saw above, either $y>0$ on $(a,0)$ or $y<0$ on $(a,0).$ Because $y'(a) =1,$ we are in the first case and $y$ is strictly increasing on $[a,0].$
Claim: $y(c)=0$ for some $c>0.$ If the claim is true, then $b= \min\{c > 0:y(c)=0\}$ is positive. Thus $y$ satisfies the DE on $[a,b],$ $y>0$ on $(a,b),$ and $y(a)=y(b)=0.$ This $y$ is therefore a counterexample to both 3. and 4.
Proof of claim: Suppose this is false. Then $y>0$ on $[0,\infty).$ This implies $y$ is strictly concave on $[0,\infty).$ Two cases: i) $\inf_{[0,\infty)} y > 0;$ i) $\inf_{[0,\infty)} y = 0.$
Case i) implies $y''(x) \to -\infty$ as $x \to \infty.$ That implies $y'(x)<0$ for large $x.$ But $y$ is concave on $[0,\infty),$ hence stays below its tangent lines there. Since there are tangent lines with negative slope, we see $y(x) < 0$ for large $x,$ contradiction.
Case ii) implies $y(x) < y(0)$ for some $x>0.$ By the MVT, $(y(x)-y(0))/x = y'(x_0)$ for some $x_0\in (0,x).$ Thus $y'(x_0) < 0,$ and again we arrive at a tangent line of $y$ with negative slope, giving $y(x) < 0$ for large $x,$ contradiction.
Thus the claim is proved, and we're done.
