I am reviewing a CIR model problem, where $r_t$ has following dynamics $$dr_t=a(b-r_t)dt+\sigma \sqrt{r_t} dW_t^* \quad \quad (1)$$ for some constants $ab>\frac{\sigma^2}{2} \quad$
Letting T be a fixed date and $f_{\lambda}$ a function defined for some constant $\lambda >0$ $$f_{\lambda}(t,r)=E^*[e^{-\lambda {r_{T}}}|r_t=r] \quad \quad (2)$$
a) derive PDE satisfied by the function $f_{\lambda}$
b) show that the function $f_{\lambda} (t,r)=e^{-A_{\lambda}(T-t)-B_{\lambda}(T-t)r_t}$ satisfies the PDE
I guess the subject function can be expressed as $$f(t,r_t)=e^{- \lambda r} \quad \quad (3)$$
My first thought was just to calculate the $df(t,r_t)$ using Ito formula and substituting for the $dr_t$ which would yield $$df(t,r_t)=f_t dt + f_r dr_t+ \frac{1}{2} f_{rr} d<r>_t=$$ $$=(f_t + a(b-r_t) f_r + \frac{1}{2} r f_{rr})dt+ \sigma \sqrt{r_t} dW_t^* \quad \quad (4)$$
Wouldn't it be already answer to a?
However the solution presents a different approach which I don't understand. It gives the Feynman-Kac equation as solution to a),
further the equation $f_{\lambda} (t,r)=e^{-A_{\lambda}(T-t)-B_{\lambda}(T-t)r_t}$ is plugged into the Feynman-Kac equation, and then using a system of 2 ODE's $A(\tau)$ $B(\tau)$ are derived.
Can anybody explain the proceedings please? I am missing "the big picture" here.
I don't understand why it starts with Feynman-Kac and why only deriving A and B proves already that they satisfy the PDE.