We consider a Feller-Dynkin Markov process $X$ with generator $G$, which, when restricted to $C^2$ functions with compact support, is given by $Gf(x) = \frac{c(x)}{2}f''(x)$, where $c$ is a positive continuous function on $\mathbb{R}$. Now let $a<x<b$, and let $\tau$ be the hitting time of $\{a,b\}$. I would like to show:
$$\mathbb{E}_x [\tau] = \int_a^b \frac{2}{c(z)} \frac{(x \wedge z - a)(b - x \vee z)}{b-a} dz$$
where $\mathbb{E}_x$ means we condition on $X_0 = x$. I do not really know what to try here. I thought about using Doob's optional stopping theorem, since I used it in another exercise once to compute the expected value of a stopping time. We need some martingale to apply it to, and we know that $$M_t := f(X_t) - \int_{0}^t Gf(X_s) ds$$ is a martingale. Now however, I have no clue how we should pick the function $f$. Also, I realized that the generate looks similar to the generator of Brownian Motion, which is $\frac{1}{2} f''$, but I don't see the resemblance here, what could I try?
In Karatzas-Shreve they compute the expected hitting time in 5.59 for general SDEs $dX=b(X_t)dt+\sigma(X_t)dB_t$
$$E_{x}[T_{a,b}]=M_{a,b}(x)=-\int^x_{a} (p(x)-p(y))m(dy)+\frac{p(x)-p(a)}{p(b)-p(a)}\int^b_{a} (p(b)-p(y))m(dy),$$
for speed measure $m(dy)=\frac{2dy}{p'(y)\sigma^{2}(y)}$ and scale function
$$p(x)=\int^x_{a}exp(-2\int^{y}_{a}\frac{b(z)}{\sigma^{2}(z)})dy. $$
Based on your generator, I am guessing $b=0$ and $\sigma^{2}(x)=c(x)$. So it is a bit more complicated since to get p(x), one needs to integrate $1/c(x)$, which is unknown here.