We want to find the differential equation satisfied by the operator $\tilde{U}(t, t_0)$, which is defined as

$$\tilde{U}(t,t_0) = \tilde{T}{\rm exp}\left[ -i \int_{t_0}^{t} {\mathcal H}(\tau ) d \tau \right].$$

Here $\tilde{T}$ is the antichronological ordering operator defined as

$$\tilde{T}\left[ A(t_1) B(t_2) \right] = \left \{ \begin{arr... ..._1 < t_2) \\ B(t_2) A(t_1) & (t_1 > t_2). \end{array} \right.$$

We examine the time evolution of $\tilde{U}(t,0)$ by comparing it with $\tilde{U}(t + \delta t, 0)$. We have,

$$\tilde{U}(t +\delta t, t_0) = \tilde{T}{\rm exp} \left[ -i \int_{t_0}^{t} {\mathcal H}(\tau ) d \tau -i \delta t {\mathcal H}(t+) \right]$$

$$= \tilde{T}\sum_n \frac{(-i)^n}{n!} \left[ \int_{t_0}^{t} {\mathcal H}(\tau ) d \tau + \delta t {\mathcal H}(t+) \right]^n$$

$$= \tilde{U}(t, t_0) + (-i) \tilde{T}\sum_{n=1}^{\infty} \frac{(-i)^... ...t_{t_0}^{t} {\mathcal H}(\tau ) d \tau \right]^{n-1} + {\mathcal O}(\delta t)^2$$

$$= \tilde{U}(t, t_0) -i \tilde{U}(t, t_0) \delta t {\mathcal H}(t) + {\mathcal O}(\delta t)^2.$$ (1)

Here $t+ = t + \epsilon$, with $\epsilon \rightarrow 0$. Due to anti time ordering, in the last line of the above expression, ${\mathcal H}(t+)$ moves to the right of all other terms. Thus, $\tilde{U}(t, t_0)$ obeys the differential equation

$$\frac{d}{dt} \tilde{U}(t, t_0) = \left( -i \right) \tilde{U}(t, t_0) {\mathcal H}(t).$$ (2)