TITLE: How was this PDE solution guessed?
QUESTION [4 upvotes]: I'm reading Steven Shreve's Stochastic Calculus for Finance, in pp 274, there's such a PDE to be solved:
$$f_t(t,r)+(a(t)-b(t)r)f_r(t,r)+\frac12  \sigma^2(t)f_{rr}(t,r)=rf(t,r)$$
with terminal condition
$$f(T,r)=1$$
for all $r$.
Then, it jumped to:

We initially guess and subsequently verify that the soltion has the
form
$$f(t,r)=e^{-rC(t,T)-A(t,T)}$$

Of course the guess is correct and the solution is found:
$$C(t,T)=\int_t^T e^{-\int_t^s b(v) dv} ds$$
$$A(t,T)=\int_t^T \left(a(s)C(s,T)-\frac12 \sigma^2(s)C^2(s,T)\right)ds$$
But I'm a bit lost here -- how did it make the first guess that the solution is of the form
$$f(t,r)=e^{-rC(t,T)-A(t,T)}$$
? Is this something usual in PDE solutions? or is there some well-known facts in PDE that suggest such a form?

REPLY [2 votes]: There are clues in the form of the equation, though a finance intuition might be a better guide. Taking $a$, $b$, and $\sigma$ all zero suggests solutions like $e^{rt}$. Putting those terms back then you see that the plain $t$ won't do, so perhaps $e^{r c(t)}$ will be a possible place to begin looking. The other clue is the common practice of looking for product solutions like $g(r)h(t)$, or maybe in this case $e^{r c(t)} h(t)$, which seems to have worked.
It is important to realize that finding a formula that works does not prove that you have found THE solution. You have to make a uniqueness argument to assert that.
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