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Let $X_o=x$, $dX_t=\frac{1}{X_t}dt+X_tdW_t$, $W_t$ is a brownian motion i am thinking of trying $Y_t=\frac{X_t^2}{2}$ and apply ito's lemma on $Y_t$

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i think $Y_t=\frac{-X^2_t}{4}$ will do the trick –  Glenn K. Nov 8 '11 at 6:25
    
As a solution of what? –  Did Nov 8 '11 at 6:27
    
i mean trying $Y_t$ , and then apply ito's lemma on $Y_t$ and find $X_t$ –  Glenn K. Nov 8 '11 at 6:28
    
You might wish to reconsider: solving an equation usually means expressing the unknown as an explicit function of the given, here the unknown is X and the given is W so you cannot express a solution as a function of X. –  Did Nov 8 '11 at 6:33
    
$W_t$ is brownian motion –  Glenn K. Nov 8 '11 at 6:35
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Let $Y_t = \mathrm{e}^{t- 2 W_t} X_t^2$. Then, applying Ito's lemma: $$ \mathrm{d} Y_t = \mathrm{e}^{t- 2 W_t} \mathrm{d} (X_t^2) + X_t^2 \mathrm{d} \mathrm{e}^{t- 2 W_t} + \mathrm{d} \mathrm{e}^{t- 2 W_t} \cdot \mathrm{d} (X_t^2) $$ Since $\mathrm{d} \mathrm{e}^{t- 2 W_t} = \mathrm{e}^{t- 2 W_t} \left( 3 \mathrm{d} t - 2 \mathrm{d} W_t \right)$ and $\mathrm{d}(X_t^2) = (X_t^2 + 2) \mathrm{d} t + 2 X_t^2 \mathrm{d} W_t$, we get: $$ \mathrm{d} Y_t = 2 \mathrm{e}^{t- 2 W_t} \mathrm{d} t $$ Also, initial condition is $Y_0 = \mathrm{e}^{0 - 2 W_0}X_0^2 = x^2$. Therefore: $$ \mathrm{e}^{t- 2 W_t} X_t^2 = Y_t = x^2 + \int_0^t 2 \mathrm{e}^{s - 2 W_s} \mathrm{d} s $$ This provides the solution: $$ X_t = \operatorname{sign}(x) \sqrt{ x^2 \mathrm{e}^{2 W_t - t} + 2 \mathrm{e}^{2 W_t - t} \int_0^t \mathrm{e}^{s - 2 W_s} \mathrm{d} s } $$

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