Been stuck on trying to solve the following SDE:
\begin{equation} dX_t = \left(X_t + t^2\right)dt + e^{W_t}\,dW_t \end{equation} Where $W_t$ represents the standard Brownian motion and $X_0 = 1$
My attempt so far
Using ito Lemma, $dX_t = \left(\frac{\partial f}{\partial t} + \frac{1}{2}\frac{\partial f^2}{\partial^2W_t}\right)dt + \frac{\partial f}{\partial W_t} dW_t$
I equated the terms giving me \begin{align} \frac{\partial f}{\partial t} + \frac{1}{2}\frac{\partial f^2}{\partial^2W_t} = X_t + t^2 &= f + r^2 \tag{1}\label{1} \\ \frac{\partial f}{\partial W_t} &= e^{-W_t} \tag{2}\label{2} \end{align} Then I used separation of variables on equation (\ref{2}) \begin{align} \int df &= \int e^{-W_t} dW_t \\ f &= -e^{-W_t} + a(t) \end{align}
Then using this equation of $f$ I found the following derivatives: \begin{align} \frac{\partial f}{\partial t} = a'(t) && \frac{\partial f}{\partial W_t} = e^{W_t} && \frac{\partial f^2}{\partial^2W_t} = -e^{-W_t} \end{align} Which I then substituted into (\ref{1}), giving: \begin{align} a'(t) +-\frac{1}{2}e^{-W_t} &= f+t^2 \quad \text{Substituting our equation for $f$}\\ a'(t) - a(t) -t^2 +\frac{1}{2}e^{-W_t} &= 0 \end{align}
From here I am not too sure how to solve $a(t)$ or even if that's what I need to find the solution. I have seen this method used in a simpler example but not sure if it applies here.
My question actually asked me to find the expectation of $X_t$ but my thinking was, that I first must find the solution to the SDE to find the expectation.
EDIT: Following from Chaos's answer (??)
\begin{align} dY_t=e^{-t}t^2dt+e^{W_t-t}dW_t \end{align}
Which solving gives:
\begin{equation*} Y_t = -e^{-t}\left(t^2+2t+2\right) + e^{W_t-t} + C_1 \end{equation*} \begin{equation*} X_t = e^{W_t} + \frac{C_2}{e^{-t}} - t^2-2t-2 \end{equation*}
Using our initial condition:
$$X_0 = 1 + C_2 - 2 = 1 \implies C_2 = 2$$
Giving:
\begin{align*} E\left[X_t\right] = E\left[e^{W_t} + \frac{2}{e^{-t}} - t^2-2t-2\right] = \int_{-\infty}^\infty \left(e^{z} + \frac{2}{e^{-t}} - t^2-2t-2\right)\mathcal{F}(z)\;\text{d}{z} \end{align*} With $\mathcal{F}(x)$ being the pdf of $\mathcal{N}(0,t)$ distribution.
Let $Y_t:=e^{-t}X_t$, and using the Itô formula compute its stochastic differential as
$$dY_t=e^{-t}dX_t-e^{-t}X_tdt$$ now replace $dX_t$ with the expression you are given.
At this point you will notice that you no longer have an SDE but rather the differential form of a Itô process, $$dY_t=e^{-t}t^2dt+e^{W_t-t}dW_t.$$
Now just integrate both sides and proceed to obtain $X_t$, from there I am sure you'll manage to obtain the expectation by yourself.