We first solve for the differential equation:
d B d t = a B − m d B = ( a B − m ) d t d B a B − m = d t ( separating the variables ) ∫ d B a B − m = ∫ d t 1 a ln | a B − m | = t + C ln | a B − m | = a t + C ( or aC, but C is arbitrary so we still call it C ) | a B − m | = e a t + C = A e a t ( where A = e C ) a B − m = ± A e a t ( removing the absolute values gives a ± ) B = 1 a ( m − A e a t ) ( A here is ± A is arbitrary and still unknown ) {\displaystyle {\begin{aligned}{\frac {dB}{dt}}&=aB-m\\dB&=(aB-m)dt\\{\frac {dB}{aB-m}}&=dt\quad ({\text{separating the variables}})\\\int {\frac {dB}{aB-m}}&=\int dt\quad \\{\frac {1}{a}}\ln |aB-m|&=t+C\\\ln |aB-m|&=at+C\quad ({\text{or aC, but C is arbitrary so we still call it C}})\\|aB-m|&=e^{at+C}=Ae^{at}\quad ({\text{where }}A=e^{C})\\aB-m&=\pm Ae^{at}\quad ({\text{removing the absolute values gives a }}\pm )\\B&={\frac {1}{a}}(m-Ae^{at})\quad ({\text{A here is }}\pm A{\text{ is arbitrary and still unknown}})\end{aligned}}}
Now we can use the initial conditions with B ( 0 ) = 30000 {\displaystyle B(0)=30000} and a = 0.02 = 1 / 50 {\displaystyle a=0.02=1/50} to find:
B ( 0 ) = 30000 = 50 ( m − A ) ⟹ A = m − 600 {\displaystyle B(0)=30000=50(m-A)\implies A=m-600} and thus
B ( t ) = 50 ( m − ( m − 600 ) e 0.02 t ) = 50 ( ( 600 − m ) e 0.02 t + m ) {\displaystyle B(t)=50(m-(m-600)e^{0.02t})=50((600-m)e^{0.02t}+m)} .
and 
to find:
and thus
.
