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

Separation of variables tells us that $\int \frac{1}{B} d B = \int (r t) d t .$ Integrating, we find $\log | B | = \frac{r t^{2}}{2} + C,$ which yields $| B | = e^{r t^{2} / 2 + C} = e^{r t^{2} / 2} \cdot e^{C} = A e^{r t^{2} / 2} .$ We can remove the absolute values at the expense of allowing $A$ to be also negative, so the general solution is $B = A e^{r t^{2} / 2} .$ To find a specific solution, we must use the initial value $B (0) = 100$ , which implies that $B (t) = 100 e^{r t^{2} / 2} .$