Differential equation and discrepency

The function is given by
\[f(x) = \int_{0}^{x^{2024}} e^{-t^2} \, dt.\]
The first derivative of \( f(x) \) using the chain rule is:
\[f'(x) = \frac{d}{dx} \left( \int_{0}^{x^{2024}} e^{-t^2} \, dt \right) = e^{-(x^{2024})^2} \cdot \frac{d}{dx} \left( x^{2024} \right) = e^{-x^{4048}} \cdot 2024 x^{2023}\]


The second derivative of $f(x) $ using the product rule is:
\[f''(x) = \frac{d}{dx} \left( 2024 x^{2023} e^{-x^{4048}} \right)\]
Hence
\[f''(x) = 2024 \left( 2023 x^{2022} e^{-x^{4048}} + x^{2023} \cdot \frac{d}{dx} (e^{-x^{4048}}) \right)\]
Using the chain rule on the exponential term:
\[\frac{d}{dx} (e^{-x^{4048}}) = e^{-x^{4048}} \cdot (-4048 x^{4047})\]
Therefore
\[f''(x) = 2024 \left( 2023 x^{2022} e^{-x^{4048}} - 4048 x^{6070} e^{-x^{4048}} \right).\]

Now consider the differential equation $ y'' + p(x)y' + q(x)y = 0 $. Substituting $ y = f(x)$, $y' = f'(x) $, and $ y'' = f''(x) $, we get:
\[f''(x) + p(x)f'(x) + q(x)f(x) = 0\]
Substituting the specific expressions for $f(x) $, $ f'(x) $, and $ f''(x) $: 
\[2024 \left( 2023 x^{2022} e^{-x^{4048}} - 4048 x^{6070} e^{-x^{4048}} \right) + p(x) \left( 2024 x^{2023} e^{-x^{4048}} \right) + q(x) \left( \int_{0}^{x^{2024}} e^{-t^2} \, dt \right) = 0.\]
Divide by $x^{2022}$ to get 
\[2024 \left( 2023  e^{-x^{4048}} - 4048 x^{4048} e^{-x^{4048}} \right) + p(x) \left( 2024 x e^{-x^{4048}} \right) + q(x) \frac{\left( \int_{0}^{x^{2024}} e^{-t^2} \, dt \right)}{x^{2022}} = 0       (*)\]
Note that 
\[\lim_{x\rightarrow 0} \frac{\left( \int_{0}^{x^{2024}} e^{-t^2} \, dt \right)}{x^{2022}} =0.\]
So letting $x\rightarrow 0$ in (*) we get 
\[2024 \left( 2023  - 0 \right) + p(x) \left( 0 \right) + q(x) 0 = 0,       (*)\]
and hence 
\[2024 \times 2023=0\]
which is a contradiction.