SDE, How to calculate and check the result

Question

Hey i need to solve this SDE

$dX_t=-X_tdt+e^{-t}dW_t, X_0=x_0\in\mathbb{R}$

I do this:

$f(t)=-1, g(t)=0, h(t)=e^{-t}$

$dX_t+X_tdt=e^{-t}dW_t$

Now i multiply by: $e^{-\int_0^t f(s)ds}=e^{-t}$

$e^{-t}dX_t-e^{-t}X_tdt=e^{-2t}dW_t$

and now the left side is a derivative $d(e^{\int_0^t f(s)ds}\cdot X_t)=d(e^{-t}X_t)$.

$d(e^{-t}X_t)=e^{-2t}dW_t$

and

$e^{-t}X_t-e^{0}X_0=\int_0^te^{-2s}dWs$

$X_t=x_0e^{t}+e^{t}\int_0^te^{-2s}dWs$ and here again i have no idea how to check that the result is okay so how to calculate $dX_t$ Can anyone help?

Answer 1

The process $X$ looks alot like the Ornstein–Uhlenbeck process.

Let us first define the following process: for all $t\geq0$, $Y_t = X_te^{t}$ with $Y_0 = x_0$. Thanks to Integration by parts, we have: \begin{align} dY_t &= d( X_te^{t}) = e^tdX_t + X_te^tdt + 0 \\ &= e^t(-X_tdt+e^{-t}dW_t) + X_te^tdt\\ &= dW_t \end{align} Thus, we have that $Y_t = x_0 + W_t$ but using the expression of $Y$, we then ended up with: \begin{equation} X_t= x_0e^{-t} + e^{-t}W_t \end{equation}