Define the quadratic variation of a semimartingale $(X_t)_{t \geq 0}$ by
$$[X,X]_t := \mathbb{P}-\lim_{n \to \infty} \sum_{j=0}^n (X_{t_j}-X_{t_{j-1}})^2$$
where $\Pi_n := \{t_0<\ldots<t_n<t\}$ is a sequence of partitions such that $|\Pi_n| \to 0$. Moreover, we set $$[X,Y]_t := \frac{1}{4} ([X+Y]_t-[X-Y]_t).$$
Using this definition it is not difficult to show the following result.
Lemma 1 Let $(M_t)_{t \geq 0}$ a continuous square-integrable martingale and $(A_t)_{t \geq 0}$ a continuous process of bounded variation. Then,
- $[M]_t$ is the unique previsible process such that $M_t^2-[M]_t$ is a martingale.
- $[M,A]_t=[A,A]_t=0$
Now we are ready to calculate the quadratic covariation. By definition,
$$X_t\pm Y_t= \underbrace{\int_0^t (2+X_s \pm Y_s) \, dB_s}_{=:M_t} + \underbrace{c_{\pm}+ \int_0^t (6+3X_s \pm 3Y_s) \, ds}_{=:A_t}.$$
where $c_+=2$, $c_-=0$. Obviously, $(M_t)_{t \geq 0}$ is a (continuous) martingale and $(A_t)_{t \geq 0}$ of bounded variation. Consequently, we obtain by applying Lemma 1
$$[X \pm Y,X \pm Y]_t = [M,M]_t+2[M,A]_t+[A,A]_t = \int_0^t (2+X_s \pm Y_s)^2 \, ds.$$
Hence,
$$[X,Y]_t = \frac{1}{4} ([X+Y]_t-[X-Y]_t) = \int_0^t (2+X_s) \cdot Y_s \, ds \tag{1}$$
In order to compute $\mathbb{E}[X,Y]_t$, we have to find $\mathbb{E}Y_t$ and $\mathbb{E}(X_t \cdot Y_t)$. Since stochastic integrals with respect to a Brownian motion are martingales, we find
$$f(t) :=\mathbb{E}(Y_t)=1+\underbrace{\mathbb{E} \left( \int_0^t Y_s \, dB_s \right)}_{0} + 3 \int_0^t \mathbb{E}(Y_s) \, ds$$
i.e. $f$ satisfies the ODE
$$f'(t) = 3f(t) \qquad f(0)=1$$
Obviously, the unique solution is given by $$\mathbb{E}Y_t = f(t)= e^{3t} \tag{2}$$ Similarly, we obtain from Itô's formula that
$$g(t) := \mathbb{E}(X_t \cdot Y_t) =1+\mathbb{E} \left( \int_0^t(7X_s Y_s+8Y_s) \, ds \right) = 8 \int_0^t f(s) \, ds + 7 \int_0^t g(s) \, ds$$
i.e. $$g'(t) = 8f(t)+7g(t) = 8e^{3t}+7g(t) \tag{3}$$
This ODE can be solved explicitely; I leave it to you. Combining $(1)$, $(2)$ and $(3)$ allows us to compute $\mathbb{E}[X,Y]_t$.
Assume that $X$ solves
$$
dX_t = \mu(t,X_t) dt + \sigma(t, X_t) dW_t.
$$
In integral form, this means
$$
X_t = X_0 + \int_0^t \mu(s,X_s) ds + \int_0^t \sigma(s,X_s) dW_s.
$$
Now, the quadratic variation of a stochastic integral process $H\cdot Y$, where $(H\cdot Y)_t = \int_0^t H_s dY_s$, is
$$
[H\cdot Y]_t = \int_0^t H_s^2 d[Y]_s.
$$
You can find this in a special case as Proposition 3.2.17 of Karatzas and Shreve's "Brownian Motion and Stochastic Calculus", or in a general version for continuous semimartingales in Section IV.31 of Rogers and William's "Diffusions, Markov processes and Martingales". Using this, we obtain (with $\mu_t = \mu(t,X_t)$ and $\sigma_t = \sigma(t, X_t)$ as convenient shorthands, and $A_t = t$)
$$
[X]_t = [\mu \cdot A]_t + [\sigma\cdot W]_t
= \int_0^t \mu_s^2 d[A]_s + \int_0^t \sigma_s^2 d[W]_s \\
= \int_0^t \sigma(s,X_s)^2 ds,
$$
where we have used $[A]_t = 0$, since all continuous processes of finite variation have zero quadratic variation, and $[W]_t = t$. As for the geometric Brownian motion, we have that $X$ is a GBM if it satisfies
$$
dX_t = \mu X_t dt + \sigma X_t dW_t,
$$
which yields
$$
[X]_t = \int_0^t \sigma^2 X_s^2 ds.
$$
In particular, $[X]$ isn't immediately seen to satisfy any SDE or ODE (as $[X]$ does not figure in the right-hand side of the above). The above is simply a representation of $[X]$ in terms of $X$, but I doubt that there in general exists and SDE or ODE satisfied by $[X]$.
Best Answer
There are several definitions for the quadratic variation: