Non existence of solution to a Differential Equation via Integral Equation

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I have the equation

$$u'' + \alpha \dfrac{g \left( t \right)}{x_0^{\alpha + 1} \left( t \right)} u = 0$$

which, in literature, is called Hill's equation. Here, $g, x_0$ are $T$-periodic and $\alpha > 0$. Also, $g$ is sign-changing with the mean value $\bar{g} = \dfrac{1}{T} \int\limits_{0}^{T} g \left( t \right) \ dt < 0$ and $x_0$ is a positive function.

In an earlier post: Conditions on a coefficient so that the differential equation has no periodic solutions, I had asked about conditions on the coefficient so that the equation of the above type has no periodic solutions other than the trivial solution. However, the condition (sufficient) is that the coefficient is non-positive. In our context, it would mean that the function $\alpha \dfrac{g \left( t \right)}{x_0^{\alpha + 1} \left( t \right)} \leq 0$ for all $t \in \left[ 0, T \right]$ (infact, whole of $\mathbb{R}$). But, since $x_0$ is positive and $\alpha > 0$, it would mean that $g$ is non-positive (which does not fit the conditions first imposed on $g$).

Therefore, I would like to find more conditions on $g$ so that the equation mentioned above has no periodic solutions. To do so, I tried to convert the equation into an integral equation. Since finding periodic solutions is a boundary value problems, it leads us to the Fredholm integral equation,

$$u \left( t \right) = u \left( 0 \right) + \int\limits_{0}^{T} K \left( t, s \right) u \left( s \right) \ ds$$

where

$$K \left( t, s \right) = \begin{cases} \dfrac{\left( t - T \right) \left( t - s \right) \alpha}{T} \dfrac{g \left( s \right)}{x_0^{\alpha + 1} \left( s \right)} & 0 \leq s \leq t \\ \dfrac{\left( t - s \right)t \alpha}{T} \dfrac{g \left( s \right)}{x_0^{\alpha + 1} \left( s \right)} & t \leq s \leq T \end{cases}$$

To incorporate the periodic conditions, we also have another equality that must be satisfied

$$\int\limits_{0}^{T} \dfrac{g \left( s \right)}{x_0^{\alpha + 1} \left( s \right)} u \left( s \right) \ ds = 0$$

Now, from here I have no clue as to what conditions should we impose on $g$ so that the integral equations (above) do not have solutions. Any help or reference for the same will be appreciated!