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Solve the SDE $d\delta_{t}=\mu\delta_{t}dt+\sigma\delta_{t}dW_{t}$ by applying the Ito formula and the final solution should be in the form of $\delta_{t}=\delta_{0}\exp\left((\mu-\frac{1}{2}\sigma^{2})t+\sigma W_{t}\right)$

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The useful form of Ito's lemma is that:

$$f(W_b,b)-f(W_a,a)=\int_a^b\partial_wf(W_t,t)dW_t+\int_a^b\left(\frac{1}{2}\partial^2_wf(W_t,t)+\partial_tf(W_t,t)\right)dt$$

which is equivalent to:

$$df=\partial_tfdt+\partial_wfdW_t+\frac{1}{2}\sigma\partial^2_wfdt$$

for a brownian motion $W_t$. You have then that $\sigma\delta_t=\partial_w\delta(W_t,t)$ and that $\mu\delta_t=\partial_t\delta(W_t,t)+\frac{1}{2}\partial^2_w\delta_t(W_t,t)$. The first equation gives $\delta(W_t,t)=C(t)\exp(\sigma W_t)$, so plugging this into the second equation and dividing by $\exp(W_t)$ gives: $\mu C(t)=C'(t)+\frac{\sigma^2}{2}C(t)$. This is again easy to solve and gives $C(t)=D\exp\left[\left(\mu-\frac{\sigma^2}{2}\right)t\right]$. Combining, we have $\delta(W_t,t)=D\exp\left((\mu-\frac{\sigma^2}{2})t+\sigma W_t\right)$. At $t=0$, $\delta(W_0,0)=\delta_0=D$, so we finally end up with:

$$\delta(W_t,t)=\delta_0\exp\left(\left(\mu-\frac{\sigma^2}{2}\right)t+\sigma W_t\right)$$

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Is there anyway that I can set up a substitution to solve it and transfer it into the final answer. –  Tony Dec 12 '12 at 5:41
    
Yes, take a look here. I've just done this in full generality: en.wikipedia.org/wiki/Geometric_Brownian_motion#Solving_the_SDE –  Alex R. Dec 12 '12 at 5:43
    
Can you give out the substitution directly? –  Tony Dec 12 '12 at 5:48
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