# solution of SDE: $dS_t=(\alpha S_t+f(t))dW_t$

does someone know how to solve the following SDE

$$dS_t=(\alpha S_t+f(t))dW_t, S_0=s$$

where $f(t)$ is a deterministic function and $W_t$ is a standard brownian motion. Is there a explicit solution for this SDE? Many thanks!

Note that the SDE $$dS_t = (\alpha \cdot S_t+f(t)) \, dW_t \qquad S_0 = s$$ is a linear SDE and therefore there is indeed an explicit formula for the solution of the SDE. One approach is the following:

1. Solve the homogeneous SDE $$dS_t = \alpha \cdot S_t \qquad X_0 = x$$ There are several possibilities to do so: Either you have some intuition how the solution might look like or you apply Itô's formula to $X_t := g(W_t)$ to obtain conditions on (the derivatives of) $g$ allowing you to determine $g$.
2. Solve the inhomogeneous SDE $$dS_t = (\alpha \cdot S_t+f(t)) \, dW_t$$ One approach is the so-called variation of constants: Let $S_t^0$ such that $\frac{1}{S_t^0}$ solves the homogeneoues SDE, $S_0^0 = 1$ (we solved this SDE in the first step, so there is an explicit formula). Now we apply Itô's formula to $$Z_t := S_t \cdot S_t^0$$ which works fine since we know $dS_t$, $dS_t^0$. Therefore, we can determine $Z_t$ and consequently also $S_t = \frac{Z_t}{S_t^0}$.

Literature: e.g. René L. Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes

It is easy to check that \begin{align*} d\left(e^{-\frac{\alpha^2}{2}t - \alpha W_t } S_t \right) = e^{-\frac{\alpha^2}{2}t - \alpha W_t } f(t) dW_t. \end{align*} Then, $S_t$ can be solved subsequently.

• What is a motivation behind considering the discounting factor? – Idonknow Dec 11 '19 at 15:39
• It is the integration factor, and is motivated by the usual treatment for mean-reverting processes. – Gordon Dec 11 '19 at 16:24
• Thanks for your reply. Is the integrating factor same as in first order ordinary differential equation? – Idonknow Dec 11 '19 at 23:39
• Yes. Very similar idea. – Gordon Dec 12 '19 at 12:19