# Solve the SDE $dX_t = \frac{1}{2 X_t} dt + dB_t$ [closed]

Solve the following stochastic differential equations

$dX_t = \frac{1}{2 X_t} dt + dB_t$

or equivalently with a transformation $Y_t = X_t^2$

$dY_t = dt + 2 \sqrt{Y_t} dB_t$ with $Y_0 = y_0 > 0$

where B is one dimentional standard Brownian motion.

The solution is Given by $Y_t = (\sqrt{y_0} + B_t)^2$. You can check this by finding the differential of the solution using Ito's formula. This solution was by guess. The way to solve this problem is by introducing a transformation of $Y_t$ that will give a linear SDE that one can solve easily. Then the solution can be found by the inverse of the used transformation.

## closed as off-topic by Did, Moishe Kohan, Davide Giraudo, Ivo Terek, Swapnil TripathiDec 18 '14 at 19:53

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• Just to be sure: It's really $\frac{1}{2Y_t}$ and not $\frac{1}{2} Y_t$ (in the 1st SDE), right? And: What have you tried? Do you know any approach to solve SDEs? – saz Dec 18 '14 at 15:34
• If one try to solve the second SDE, we can let $X_t = \sqrt{Y_t}$. Then $dX_t = 0.5 Y_t^{-0.5} dY_t = \frac{1}{2 X_t} dY_t$ and we get the first one. I'm learning SDEs. I usually use Ito formula. – user144410 Dec 18 '14 at 15:37
• As far as I see you didn't apply Itô's formula correctly. However; this means that you actually want to solve the second SDE - am I right? – saz Dec 18 '14 at 15:40
• Here I did not apply any Ito formula, I was answering you regarding "Do you know how to approach SDEs. Both SDEs are the same up to the transformation $X_t = \sqrt{Y_t}$. A solution to any of them will yield the solution to the other. – user144410 Dec 18 '14 at 15:42
• I think the way to solve this is to find a transformation that gives a linear SDE. – user144410 Dec 18 '14 at 15:44

Hint: For the second SDE use the substitution $$Z_t := \exp(-\sqrt{Y_t}).$$ For the general approach see e.g. (the second part of) this answer.
Remark: As @user144410 pointed out, the solution of the first SDE can be obtained from the second one using the transformation $X_t = \sqrt{Y_t}$.
• I applied Itô's formula, i.e. calculated $dZ_t$. – saz Dec 18 '14 at 19:32