Solving SDE: $dX(t) = udt + \sigma X(t)dB(t)$ Solve the SDE: $dX(t) = udt + \sigma X(t)dB(t)$
Provided Question
The SDE is $dX(t) = udt + \sigma X(t)dB(t)$. Find $X(t)$, where $X(t)$ is some stochastic process and $B(t)$ is a Wiener process. Both $u$ and $\sigma$ are constants.
Hint: Multiply both sides by the "integrating factor" $e^{-\sigma B(t) + \frac12 \sigma^2 t}$.
Current Progress
Multiplying both sides by the appropriate integrating factor:
$$
\exp{\big( -\sigma B(t) + \frac12 \sigma^2t\big)}dX(t) = \exp{\big(-\sigma B(t)+\frac12\sigma^2t\big)}(udt + \sigma X(t)dB(t))
$$
Then set $f(t,x,b):=\exp{\big( -\sigma B(t) + \frac12 \sigma^2t\big)}X(t)$ and apply Ito's formula. Some of the required results before actually applying Ito's formula:
$
\frac{df}{dt} = \frac12\sigma^2 X(t)e^{-\sigma B(t) + \frac12 \sigma^2t} \\
\frac{df}{dx} = e^{-\sigma B(t) + \frac12 \sigma^2t} \\
\frac{df}{db} = -\sigma X(t)e^{-\sigma B(t) + \frac12 \sigma^2t} \\
\frac{d^2f}{dx^2} = 0 \\
\frac{d^2f}{db^2} = \sigma^2 X(t) e^{-\sigma B(t) + \frac12 \sigma^2t} \\
\frac{d^2f}{dxdb} = -\sigma e^{-\sigma B(t) + \frac12 \sigma^2t} 
$
Given this, we need to know the following derivatives of the quadratic variations and co-variations:
$
d\langle B\rangle_t = dt \\
d\langle X,B\rangle_t = ???
$
My Request
Please instruct me on what $d\langle X,B\rangle_t$ is equal to so that I may progress further with this problem.
 A: As mike already wrote: $\langle X,B\rangle_t = \sigma \cdot X_t \, dt$. This fact follows straight from the definition of $\langle\cdot,\cdot\rangle$.
Another approach to solve the SDE:

*

*Solve the homogeneous SDE $dX_t = \sigma \cdot X_t \, dB_t$. Idea: Set $Z_t := \log X_t$ and apply Itô's formula to $f(x) := \log x$, then $$\begin{align*} dZ_t &= \frac{1}{X_t} \, dX_t - \frac{1}{2} \frac{1}{(X_t)^2} (dX_t)^2 \\ &= \left(- \frac{1}{2} \sigma^2 \right) \, dt + \sigma dB_t \\
\Rightarrow X_t &= X_0 \cdot \exp \left(- \int_0^t \frac{1}{2} \sigma^2 \, ds \right) \cdot \exp \left(\int_0^t \sigma \, dB_s \right) \\ &= X_0 \cdot \exp \left(- \frac{1}{2} \sigma^2 \cdot t + \sigma \cdot B_t \right) \end{align*}$$

*Solve the inhomogeneous SDE: Let
$$X_t^0 := \exp  \left( \frac{1}{2} \sigma^2 \cdot t - \sigma \cdot B_t \right)$$
then we know that $\frac{1}{X_t^0}$ solves the homogeneous equation (and $X_0^0=1$).  Now set $Z_t := X_t \cdot X_t^0$ and apply Itô's formula to $f(x,y) = x \cdot y$, then
$$dZ_t = \ldots = u \cdot X_t^0 \, dt$$ thus
$$X_t = \frac{Z_t}{X_t^0} = \frac{u \cdot \int_0^t \exp \left( \frac{1}{2} \sigma^2 \cdot s - \sigma \cdot B_s\right) \, ds}{\exp \left( \frac{1}{2} \sigma^2 \cdot t - \sigma \cdot B_t\right)}$$
This approach also works in a similar fashion for the (more general) SDE
$$dX_t = (a(t)+b(t) \cdot X_t) \, dt + (c(t)+d(t) \cdot X_t) \, dB_t$$
