I have the following question which makes sense when taking into account that $X_0=0$, but I don't get the same result if we use the variable.

QUESTION: Consider the SDE $$dX_t=\alpha X_t dt+2dW_t,\quad X_0=0,$$ where $W_t$ is a standard Brownian motion and $\alpha\in\mathbb{R}$ is a constant.

Derive the unique solution.

PROFESSOR'S ANSWER: $X_t=e^{\alpha t}X_0+2e^{\alpha t}\int_0^t e^{-\alpha s}dW_s$


I first proved using one of my theorems in my notes that this SDE has a unique solution.

We have $$|\alpha x-\alpha y|+|2-2|=|\alpha x-\alpha y|=|\alpha(x-y)|\le\alpha|x-y|,$$ for all $x,y\in\mathbb{R}$. Hence, the SDE has a unique solution.

Now, consider $Z_t=e^{-\alpha t}X_t$, then Ito's formula gives: $$dZ_t=\left(-\alpha e^{-\alpha t}X_t+\alpha e^{-\alpha t}X_t\right)dt+2e^{-\alpha t}dW_t=2e^{-\alpha t}dW_t$$ THIS IS WHERE I GET CONFUSED - have I done something wrong, or is my answer the correct one?

We have $Z_t=e^{-\alpha t}X_t$, so $X_t=e^{\alpha t}Z_t$. Consider $dX_t=e^{\alpha t}dZ_t$, then integrating on both sides gives: $$\int_0^t dX_s=e^{\alpha t}\int_0^t dZ_s\implies \int_0^t dX_s=2e^{\alpha t}\int_0^t e^{-\alpha s}dW_s\implies X_t-X_0=2e^{\alpha t}\int_0^t e^{-\alpha s}dW_s\implies X_t=X_0+2e^{\alpha t}\int_0^t e^{-\alpha s}dW_s$$

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    $\begingroup$ Look at this math.stackexchange.com/questions/755396/… $\endgroup$ Aug 7, 2015 at 21:07
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    $\begingroup$ There are a lof of answers on this topic; just search for "linear SDE" or have a look at the questions with the "SDE"-tag. $\endgroup$
    – saz
    Aug 7, 2015 at 21:13

1 Answer 1


So after you used Ito's formula, you have $dZ_t = d(e^{-\alpha t}X_t) = 2e^{-\alpha t}dW_t$. Then you have $$X(t)e^{-\alpha t} - X_0 = \int_0^t 2e^{-\alpha s}dW_s$$

Thus, rearranging, you $$X(t) = X_0e^{\alpha t}+ 2e^{\alpha t}\int_0^t e^{-\alpha s}dW_s$$

which gives the professor's answer. I'm not sure which part you're confused about.


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