# Finding a solution to the SDE of $dX_t = -2 (1-t)^{-1}X_tdt + \sqrt{2t(1-t)} dW_t$.

I am trying to find the solutions to the SDE: The solution of the following SDE $$dX_t = -2 \frac{X_t}{1-t} dt + \sqrt{2t(1-t)} dW_t, \quad X_0 = 0$$

where $W_t$ is a Wiener process. I know that due to the form of the SDE above, it should be a Gaussian process. I am trying to compute the solution to $X_t$. It is my understanding that that a Gaussian process can be written as:

$$X_t = a(t)\left[x_0 + \int_0^tb(s)dB_s\right]$$

where the differential is:

$$a'(t)\left[x_0 + \int_0^tb(s)dB_s\right]dt + a(t)b(t)dB_t$$

Matching coefficients, we have that $\frac{a'(t)}{a(t)} = -2(1-t)^{-1}$ and that $a(t)b(t) = \sqrt{2t(1-t)}$. However, the solution to $a(t)$ is given to be $a(t) = K(t-1)^2$. With the initial condition that $X_0 = 0$, it seems there are multiple solutions. Would anyone have any idea how to proceed? thanks.

Since the initial condition is $X(0) = 0$ you can try the test function $$X(t) = a(t)\int_{0}^{t}b(s)dB(s),$$ and computing $dX(t)$ implies that you obtain the following ODE, which can be solved by integrating both sides \begin{align} &\frac{a'(t)}{a(t)}dt = \frac{-2dt}{1-t} \\ &\Leftrightarrow \ln|a(t)| = 2 \ln|1-t| + C \\ &\Leftrightarrow a(t) = e^{C}(1-t)^2. \end{align} Setting $e^{C} = K$ yields the answer for $a(t)$.
• Thanks, just a problem I have, I am not sure how to get $K = e^{C}$, since if we use the initial condition, $X(0) = 0$, we have that the integral part will necessarily be $0$, but it is ambiguous what $a(t)$ can be. Do you have any ideas? Thanks! Commented Apr 12, 2016 at 0:57