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I have a question:

Let $(B_t)_{t\geq 0}$ be a brownian motion. Consider the following function $u(x)$ defined by

$$u(x)=E_x\left[\exp\left(-\frac{\theta^2}{2}T_0+\int_{0}^{T_0}f(B_u)du\right)\right]$$

where $E_x$ implies $B_0=x$, $T_0=\inf\{t\geq 0: B_t=0\}$ and $f$ is a borelian bounded function. Prove that

$$\frac{u''}{2}=(\frac{\theta^2}{2}+f)u,\ \ u(0)=1$$

Does someone have an idea? Many thanks!

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