ODE problem $\frac{dB}{B(a+\frac{\sigma^2}{2} B)}=-dt$ I am analyzing solution to a problem where one of the steps was solving the below ODE, I have the final form of this equation but I cannot follow the transitions which were done.

Can anybody explain how this equation was solved?

$$\frac{dB}{B(a+\frac{\sigma^2}{2} B)}=-dt$$

which should result in
$$ \frac{B}{a+\frac{\sigma^2}{2} B}=ce^{-at}$$
Using Partial Fractions,
$$ \frac{1}{B(a+\frac{\sigma^2B}{2})} = \frac{A}{B} + \frac{C}{a+\frac{\sigma^2B}{2}} $$
$$ \therefore Aa = 1, A = \frac{1}{a}  $$
$$ A\frac{\sigma^2}{2} + C = 0, C = -A\frac{\sigma^2}{2} = -\frac{\sigma^2}{2a}  $$
$$ \frac{1}{B(a+\frac{\sigma^2B}{2})} = \frac{1}{aB} + \frac{-\sigma^2}{(a+\frac{\sigma^2B}{2})2a} $$
$$ \int\frac{1}{B(a+\frac{\sigma^2B}{2})} dB = \int \frac{1}{aB} + \frac{-\sigma^2}{(a+\frac{\sigma^2B}{2})2a} dB = \frac{\ln{B}}{a} -\frac{\ln(a+\frac{\sigma^2B}{2})}{a} $$
$$ \int\frac{1}{B(a+\frac{\sigma^2B}{2})} dB = \frac{1}{a}\ln(\frac{B}{a+\frac{\sigma^2B}{2}}) = \int -dt = -t + k $$
where $k$ is an integration constant
$$ \therefore \frac{B}{a+\frac{\sigma^2B}{2}} = e^{-at}.e^{ak} = ce^{-at} $$
where $c = e^{ak}$ is some constant.
We have: $$\frac{dB}{B(a+\frac{\sigma^2}{2} B)}=-dt$$

Grouping: $$\frac{dt}{dB}=-\frac{1}{B(a+\frac{\sigma^2}{2}B)}$$

Then use partial fraction and integrate both sides.