1. **Problem statement:** We have the differential equation $\frac{dy}{dt} = ky$, where $k$ is a constant and $t$ is time in years. The population doubles every 10 years. We need to find the value of $k$. 2. **Formula and explanation:** The general solution to $\frac{dy}{dt} = ky$ is an exponential function: $$y = y_0 e^{kt}$$ where $y_0$ is the initial population. 3. Since the population doubles every 10 years, we have: $$y(10) = 2y_0$$ Substitute into the solution: $$2y_0 = y_0 e^{k \cdot 10}$$ 4. Divide both sides by $y_0$ (assuming $y_0 \neq 0$): $$2 = e^{10k}$$ 5. Take the natural logarithm of both sides: $$\ln(2) = 10k$$ 6. Solve for $k$: $$k = \frac{\ln(2)}{10}$$ 7. Calculate the numerical value: $$k \approx \frac{0.6931}{10} = 0.06931$$ 8. **Answer:** The value of $k$ is approximately $0.069$, which corresponds to option A. This means the population grows at a rate of about 6.9% per year to double every 10 years.