Science:Math Exam Resources/Courses/MATH101 C/April 2024/Question 15/Solution 1

Separation of variables tells us that $\int {\frac {1}{B}}\mathrm {d} B=\int (rt)\mathrm {d} t.$ Integrating, we find $\log |B|={\frac {rt^{2}}{2}}+C,$ which yields $|B|=e^{rt^{2}/2+C}=e^{rt^{2}/2}\cdot e^{C}=Ae^{rt^{2}/2}.$ We can remove the absolute values at the expense of allowing $A$ to be also negative, so the general solution is

$$B=Ae^{rt^{2}/2}.$$

To find a specific solution, we must use the initial value $B(0)=100$, which implies that

$$B(t)=100e^{rt^{2}/2}.$$