Density of set of zeroes

Let $g:(\alpha,\beta)\rightarrow\mathbb{R}$ be a twice differentiable function such that $$g(x)=\frac{x^2g''(x)+xg'(x)}{x^2+1},$$ and $0<\alpha<\beta$. Assuming, $g(\alpha)=g(\beta)=0$, is it the case that $g(x)=0$ for all $x\in(\alpha,\beta)$?

For the sake of contradiction, I assume there exists $\alpha<x<\beta$ such that $g(x)>0$. By mean value, there exist $\alpha<x_1<x<x_2<\beta$ and $g'(x_2)<0<g'(x_1)$. By intermediate value, there exists $x_3\in(x_1,x_2)$ such that $g'(x_3)=0$. So $$g(x_3)(x_3^2+1)=x_3^2g''(x_3).$$ I want to use the sign of $g''(x_3), g(x_3)$ to draw conclusions about $x_3$. But if one of $g(x_3),g''(x_3)$ is zero, then other is zero too. We have no way to guarantee that both are nonzero. As a result this approach seems dead.

Any other strategy I can try?

Edit: I think this conjecture is not right. For example, what could go wrong if $g$ is oscillating on the points between $\alpha$ and $\beta$?

Let $g$ denote a continuous, second differentiable function satisfying

$$ (x^2+1)g(x)=x^2g^{\prime\prime}(x)+xg^\prime(x) $$

Furthermore, let it be given that there is an $\alpha<\beta$ such that $g(\alpha)=g(\beta)=0$.

Assume there is an $x$ such that

1. $\alpha<x<\beta$ and

2. $g(x)>0$

Then $g$ has a maximum value $g(\xi)$ at some $x^\prime\in(\alpha)$ with $g^\prime(\xi)=0$, $g^{\prime\prime}(\xi)<0$ and $g(\xi)\ge g(x)$.

Thus $(\xi^2+1)g(\xi)=\xi^2g^{\prime\prime}(\xi)$

But the left side of this equation is positive and the right side is negative, which is a contradiction.

If we assume that there is an $x\in(\alpha,\beta)$ for which $g(x)<0$ we get an analogous contradition.

Thus $g(x)=0$ for all $x\in(\alpha,\beta)$.