How to solve this simple stochastic differential equation (SDE) $dX_t=-\frac{X_t}{1-t}dt+dB_t$? (hint included)

Question

Solve the equation $dX_t=-\frac{X_t}{1-t}dt+dB_t$, $X_0=0$ in $[0,1)$. Hint: Apply the Ito folmula to $Y_t=X_t/(1-t)$. How to apply the Ito formula step by step, and how to confirm the solution? $B_t$ is a typical 1-d Brownian motion.

Answer 1

Let $f(x,t) = \frac{x}{1-t}$. Then $\partial_t f=\frac{x}{(1-t)^2}$, $\partial_x f= \frac{1}{1-t}$, $\partial _{xx} f =0$. So the second order terms in Itô's formula all vanish. We therefore have, for the $C^2$ function $f : \mathbb{R}^2 \to \mathbb{R}$, that $Y_t$ is a continuous semimartingale with decomposition: \begin{equation} Y_t = Y_0 + \int_{0}^{t} \dfrac{\partial f}{\partial t}(X_s,s) ds+ \int_{0}^{t} \dfrac{\partial f}{\partial x}(X_s,s) dX_s = \\ = \int_{0}^{t} \dfrac{X_s}{(1-s)^2} ds+ \int_{0}^{t} \dfrac{1}{1-s} \left(- \dfrac{X_s}{1-s} ds + dB_s \right) = \int_{0}^{t} \frac{dB_s}{1-s}. \end{equation} Therefore, \begin{equation} X_t = \int_{0}^{t} \dfrac{1-t}{1-s}dB_s. \end{equation} This integral with respect to Brownian motion is as nice as the answer can get - you were on the right track.

Answer 2

In case you are curious why the hint suggests $Y_t = \frac{X_t}{1-t},$ this is because the SDE given resembles the first-order linear ordinary differential equation: Rearranging the SDE gives $$dX_t + \frac{1}{1-t} X_tdt = dB_t.$$ The integrating factor is $$I = exp\left( \frac{1}{1-t} dt \right) = exp\left( - \ln |1-t| \right) = \frac{1}{1-t}.$$ Then the rest follows what @spetrevski suggested in their solution.