Science:Math Exam Resources/Courses/MATH105/April 2014/Question 05 (a)/Solution 1

We first solve for the differential equation:

$\begin{array}{rl}\frac{dB}{dt}& =aB-m\\ dB& =(aB-m)dt\\ \frac{dB}{aB-m}& =dt(\text{separating the variables})\\ \int \frac{dB}{aB-m}& =\int dt\\ \frac{1}{a}\ln |aB-m|& =t+C\\ \ln |aB-m|& =at+C(\text{or aC, but C is arbitrary so we still call it C})\\ |aB-m|& =e^{at+C}=A{e}^{at}(\text{where }A=e^C)\\ aB-m& =\pm A{e}^{at}(\text{removing the absolute values gives a }\pm )\\ B& =\frac{1}{a}(m-A{e}^{at})(\text{A here is }\pm A\text{ is arbitrary and still unknown})\end{array}$

Now we can use the initial conditions with $B(0)=30000$ and $a=0.02=1/50$ to find:

$B(0)=30000=50(m-A)⟹A=m-600$ and thus

$B(t)=50(m-(m-600)e^{0.02t})=50((600-m)e^{0.02t}+m)$.