# Solution of SDE (regarding integrating factor)

I can't seem to find the solution of the following SDE:

$$dX_t = X_t^4dt + 2X_tdW_t \\ X_0 = \beta$$

where $$W_t$$ is a Wiener process. The question gives me a hint that I should consider an integrating factor of the form $$I_t = e^{\int_{0}^{t}c(s)dW_s-\frac{1}{2}\int_{0}^{t} [c(s)]^2 ds}$$. I know that I should proceed by applying Itô's Formula to $$d(X_tI_t)$$, but I can't seem to know how to deal with the exponential. Shall I treat it as a different process?

Thanks in advance for any help given.

• If you want to follow the hint, you need Itô's formula for Itô processes; more precisely you need to compute the stochastic differential $d(X_t Y_t)$ for two Itô processes $X$, $Y$. Then put $Y=I$. – saz Jun 29 at 20:25
• Sorry, I meant formula there. Thanks a lot for the explanations. – Leonardo Lima Jun 29 at 20:57

Assume that $$(X_t)_{t \geq 0}$$ is a solution to the SDE, and set $$Z_t := X_t^{\delta}$$ for some $$\delta$$ which we will choose in a minute. By Itô's formula,

\begin{align*} Z_t-Z_0 &= \delta \int_0^t X_s^{\delta-1} \, dX_s + \frac{\delta(\delta-1)}2 \int_0^t X_s^{\delta-2} \, d\langle X \rangle_s \\ &= 2\delta \int_0^t X_s^{\delta} \, dW_s + \int_0^t \left(\delta X_s^{\delta+3} + 2 \delta (\delta-1) X_s^{\delta} \right) \, ds. \end{align*}

For $$\delta=-3$$ this becomes

$$Z_t-Z_0 = -6 \int_0^t Z_s \, dW_s + \int_0^t (24 Z_s -3) \, ds. \tag{1}$$

Since $$(1)$$ is a linear SDE, there is a general formula for its solution:

$$Z_t = \exp(6t - 6W_t) \left( Z_0 -3 \int_0^t \exp \left[-6s+6W_s \right] \, ds \right)$$

and so

$$X_t = \frac{1}{Z_t^{1/3}}$$

is a candidate for the solution of the original SDE. It remains to check using Itô's formula that it is indeed a solution.

Remark: See this question to get an idea how to come up with the transform $$Z_t = X_t^{\delta}$$.