1. **Problem Statement:** Given the differential equation $$\frac{dy}{dt} = ky$$ where $k$ is a constant and $t$ is in years, and the bacteria doubles every 5 days, find the value of $k$. 2. **Understanding the problem:** The equation describes exponential growth. The solution to this differential equation is $$y = y_0 e^{kt}$$ where $y_0$ is the initial amount. 3. **Doubling condition:** The bacteria doubles every 5 days. Convert 5 days to years: $$5 \text{ days} = \frac{5}{365} \text{ years}$$ 4. **Apply doubling condition:** When $t = \frac{5}{365}$ years, $$y = 2 y_0$$. 5. **Use the solution formula:** $$2 y_0 = y_0 e^{k \cdot \frac{5}{365}}$$ 6. **Simplify:** Divide both sides by $y_0$ (assuming $y_0 \neq 0$): $$2 = e^{k \cdot \frac{5}{365}}$$ 7. **Take natural logarithm:** $$\ln 2 = k \cdot \frac{5}{365}$$ 8. **Solve for $k$:** $$k = \frac{365}{5} \ln 2 = 73 \ln 2$$ 9. **Calculate numerical value:** Using $\ln 2 \approx 0.6931$, $$k \approx 73 \times 0.6931 = 50.6$$ (approximate to one decimal place). **Final answer:** $$k \approx 50.6$$ per year. This means the growth rate constant $k$ is approximately 50.6 when time is measured in years and the bacteria doubles every 5 days.