The rate of change of \( y \) is directly proportional to \( y \). That is: \[ \frac{d y}{d t}=k y \] If \( y=900 \) when \( t=0 \) and \( y=500 \) when \( t=10 \), then find \( y \) when \( t=25 \). Do not round the value of \( k \), but round your final answer to the nearest whole number. \[ y=[?] \]

To solve the problem, we start with the differential equation: \[ \frac{d y}{d t} = k y \] This can be rewritten and solved using separation of variables. We can integrate both sides to find: \[ \int \frac{1}{y} \, dy = \int k \, dt \] This gives us: \[ \ln |y| = kt + C \] Exponentiating both sides, we have: \[ y = e^{kt + C} = e^C e^{kt} \] Let \( C_1 = e^C \), thus: \[ y = C_1 e^{kt} \] To find \( C_1 \), we use the initial condition \( y(0) = 900 \): \[ 900 = C_1 e^{k \cdot 0} \implies C_1 = 900 \] So our equation now looks like: \[ y = 900 e^{kt} \] Next, we use the condition \( y(10) = 500 \) to find \( k \): \[ 500 = 900 e^{10k} \] Dividing both sides by 900: \[ \frac{500}{900} = e^{10k} \implies \frac{5}{9} = e^{10k} \] Taking the natural logarithm of both sides: \[ \ln\left(\frac{5}{9}\right) = 10k \implies k = \frac{1}{10} \ln\left(\frac{5}{9}\right) \] Now we want to find \( y \) when \( t = 25 \): \[ y = 900 e^{kt} = 900 e^{25k} = 900 e^{25 \cdot \frac{1}{10} \ln\left(\frac{5}{9}\right)} \] This simplifies to: \[ y = 900 e^{\frac{5}{2} \ln\left(\frac{5}{9}\right)} = 900 \left(\frac{5}{9}\right)^{\frac{5}{2}} = 900 \cdot \frac{5^5}{9^5} = 900 \cdot \frac{3125}{59049} \] Calculating this gives: \[ y \approx 900 \cdot 0.0525 \approx 47.25 \] Thus, rounding to the nearest whole number, we find: \[ y \approx 47 \] So, the final answer is: \[ y = [47] \]