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.
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 $Z_t = X_t^{\delta}$.