Consider the Stochastic Differential Equation $dX_t = a X_t dt + \sigma d W_t$ for $t \geq 0$, where $a,\sigma \in \mathbb{R}$ and $W$ denotes a Brownian Motion.
I want to find a solution of this.
As a hint I am asked to consider the ansatz $X_t = C_t e^{at}$, where (I think) $C_t = g(W_t)$ for some function $g$.
I think I need to use Ito's formula on $f:(t,W_t) \mapsto g(W_t) e^{at}$. Using that $\langle Id,Id\rangle_t = 0, \langle Id,W\rangle_t =0$ and $\langle W,W\rangle_t = t $, Ito's formula for this in the two dimensional case yields that
$$
g(W_t)e^{at} = g(W_0) + a\int_0^t g(W_s) e^{as}ds + \int_0^t g'(W_s) e^{as}dW_s +\frac{1}{2} \int_0^t g''(W_s) e^{as} ds
$$
The $g(W_0)$ term doesn't matter and as for the second term this already looks good for the equation. Hence we have to figure out how
$$
\int_0^t g'(W_s) e^{as}dW_s +\frac{1}{2} \int_0^t g''(W_s) e^{as} ds \overset{!}{=}\int_0^t \sigma d W_s = \sigma W_t
$$
I can't see how to make equal this constant $\sigma$. Do we want something like $g(t)=\sigma W_t$ with second derivative 0?
Any help is appreciated!
Best Answer
Your assumption is wrong, $C_t$ does not only depend on the point-value of $W_t$, but on the full path from $0$ to $t$. Use the Ito formula for $C_t=f(t,X_t)=e^{-at}X_t$ to get $$ dC_t=f_t\,dt+f_x\,dX_t+\frac12f_{xx}d⟨X⟩_t \\=-ae^{-at}X_t\,dt+e^{-at}(aX_t\,dt+σdW_t)+0 =σe^{-at}dW_t $$ Thus $$ C_t=C_0+\int_0^tσe^{-as}dW_s $$