I am a little confused, an exercise by a teacher has been set which says:
For $X_t = 2e^{B_t}$ Where $B_t$ is brownian motion at time $t$.
a) Find the stochastic differential $d(X_t)$
b) Find the stochastic differential equation for $X_t$
I have completed the first exercise, but am confused as to what the difference between the two is?
for a)
$$d(2e^{B_t}) = 2e^{B_t}dB_t + e^{B_t} dt$$
The not-so-exact calculation goes as $$ d(2e^{B_t})=2e^{B_{t+dt}}-2e^{B_t}=2e^{B_t}(e^{dB_t}-1)\\ =2e^{B_t}\Bigl(1+dB_t+\tfrac12(dB_t)^2+O((dB_t)^3)-1\Bigr) $$ and using the rules $dB_t^2=dt$ and to ignore all higher powers results in $$ d(2e^{B_t})=2e^{B_t}(dB_t+\tfrac12 dt)=e^{B_t}dt+2e^{B_t}dB_t $$ as you also computed, probably using the Ito formula.
Using $X_t$ to get rid of the exponentials gives the SDE $$ dX_t=X_t(\tfrac12 dt+dB_t) $$