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I'd like to solve the Langevin SDE $$dX(t)=-bX(t)dt+\sigma dW(t),\\X(0)=X_0,$$ $W(t)$ being a standard Brownian motion, using the Itô formula $$du(t,X(t)) = \frac{\partial u}{\partial t}dt + \frac{\partial u}{\partial x}dX+\frac{1}{2}\frac{\partial^2 u}{\partial x^2}\sigma^2 dt.$$ However, I don't know where to start: if I try to divide by $X(t)$ in the fist formula and integrate I find $$\frac{dX(t)}{X(t)} = -bdt+\frac{\sigma}{X(t)}dW(t) \Longleftrightarrow \ln(X(t)) = -bt+\sigma\int_0^t\frac{1}{X(s)}dW(s).$$ I could apply Itô on $\ln(X(t))$ but I wouldn't still know the right hand side.

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Apply Ito to $$ X_t\mathrm{e}^{bt}\implies \mathrm{e}^{bt}dX_t + b\mathrm{e}^{bt}X_tdt $$ so multiply your original SDE by $\mathrm{e}^{bt}$ we find $$ \mathrm{e}^{bt}dX_t = -b\mathrm{e}^{bt}X_tdt+\sigma \mathrm{e}^{bt}dW_t $$ rearrange and using the first equation I derived $$ d\left(X_t\mathrm{e}^{bt}\right) = \sigma \mathrm{e}^{bt}dW_t $$ then integrate we find $$ X_t\mathrm{e}^{bt} = X_0 +\int_0^t\sigma \mathrm{e}^{bs}dW_s $$ or (and using your notation) $$ X(t) = X(0)\mathrm{e}^{-bt} + \sigma\int_0^t\mathrm{e}^{b(s-t)}dW(s) $$

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  • $\begingroup$ No worries. :)! $\endgroup$ – Chinny84 Jun 8 '15 at 13:40

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