I am currently researching the specific image generation problem in this paper 'Score_Based Generative modeling through stochastic differential equation' At the end of page 14, the authors are using equation (5.51) in Sarkka & Solin (2019) to govern the evolution of the variance. However, the equation provided in the paper (1) seems slightly different from the equation (2) in Sarkka & Solin (2019) and since I'm relatively new to SDE, I don't understand why there's this discrepancy.
The bolded characters are the matrices
Equation in paper:
$\frac{\mathrm{d} \boldsymbol{\Sigma}_{\mathrm{VP}}(t)}{\mathrm{d} t}=\beta(t)\left(\mathbf{I}-\boldsymbol{\Sigma}_{\mathrm{VP}}(t)\right) (1)$
Equation in Sarkka & Solin (2019):
$\begin{aligned} \frac{\mathrm{d} \mathbf{P}}{\mathrm{d} t} = \mathrm{E}\left[\mathbf{f}(\mathbf{x}, t)(\mathbf{x}-\mathbf{m})^{\top}\right]+\mathrm{E}\left[(\mathbf{x}-\mathbf{m}) \mathbf{f}^{\top}(\mathbf{x}, t)\right] +\mathrm{E}\left[\mathbf{L}(\mathbf{x}, t) \mathbf{Q} \mathbf{L}^{\top}(\mathbf{x}, t)\right], \end{aligned} (2)$
The solution to equation (1) is published in the paper:
$\boldsymbol{\Sigma}_{\mathrm{VP}}(t)=\mathbf{I}+e^{\int_0^t-\beta(s) \mathrm{d} s}\left(\boldsymbol{\Sigma}_{\mathrm{VP}}(0)-\mathbf{I}\right) (3) $
Additionally, I've attempted to solve equation (1), but my result doesn't match the equation (3) presented in the paper:
$\begin{gathered} e^{\int \beta(s) d s} \frac{d \Sigma_{V P}(t)}{d t}+e^{\int \beta(s) d s} \beta(t) \Sigma_{V P}(t)=e^{\int \beta(s) d s} \beta(t) . I \\ d\left(e^{\int \beta(s) d s} \Sigma_{V P}(t)\right)=e^{\int \beta(s) d s} \beta(t) . I \\ \int_0^t d\left(e^{\int \beta(s) d s} \Sigma_{V P}(t)\right)=\int_0^t e^{\int \beta(s) d s} \beta(\tau) \cdot I d \tau \\ \Sigma_{V P}(t)-\Sigma_{V P}(0)=e^{-\int \beta(s) d s}\left(\int_0^t e^{\int \beta(s) d s} \beta(\tau) \cdot I d \tau\right) \\ \Sigma_{V P}(t)=e^{-\int \beta(s) d s}\left(\int_0^t e^{\int \beta(s) d s} \beta(\tau) \cdot I d \tau\right)+\Sigma_{V P}(0) \end{gathered}$
I really need assistance because I am currently conducting this research without any guidance or support from anyone.
You are close, but have a few mistakes.
$$\begin{gathered} e^{\int_{\color{red}{0}}^{\color{red}{t}} \beta(s) d s} \frac{d \Sigma_{V P}(t)}{d t}+e^{\int_{\color{red}{0}}^{\color{red}{t}} \beta(s) d s} \beta(t) \Sigma_{V P}(t)=e^{\int_{\color{red}{0}}^{\color{red}{t}} \beta(s) d s} \beta(t) \cdot I \\ d\left(e^{\int_{\color{red}{0}}^{\color{red}{t}} \beta(s) d s} \Sigma_{V P}(t)\right)\color{red}{/dt} =e^{\int_{\color{red}{0}}^{\color{red}{t}} \beta(s) d s} \beta(t) \cdot I \\ \int_0^t d\left(e^{\int_{\color{red}{0}}^{\color{red}{\tau}} \beta(s) d s} \Sigma_{V P}(\color{red}{\tau})\right)=\int_0^t e^{\int_{\color{red}{0}}^{\color{red}{\tau}} \beta(s) d s} \beta(\tau) \cdot I d \tau \\ \color{red}{e^{\int_0^\tau \beta(s) d s} \Sigma_{V P}(\tau) \left. \right|_{\tau=0}^t = \int_0^t \frac{d \left(e^{\int_0^\tau \beta(s) d s}\right)}{d \tau} \cdot I d \tau} \\ \color{red}{e^{\int_0^t \beta(s) d s} \Sigma_{V P}(t) - \Sigma_{V P}(0) = e^{\int_0^\tau \beta(s) d s} \left. \right|_{\tau=0}^t \cdot I} \end{gathered}$$
Can you take it from here?