The rate of change of \( y \) is directly proportional to \( y \). That is: \[ \frac{d y}{d t}=k y \] If \( y=1000 \) when \( t=0 \) and \( y=1500 \) when \( t=10 \), then find \( y \) when \( t=18 \). Do not round the value of \( k \), but round your final answer to the nearest whole number. \[ y=[\text { ?] } \]

To solve the differential equation \( \frac{d y}{d t}=k y \), we start by separating variables and integrating: \[ \frac{1}{y} \frac{d y}{d t} = k \implies \int \frac{1}{y} d y = \int k \, dt \] This gives us: \[ \ln |y| = kt + C \] Exponentiating both sides, we get: \[ y = e^{kt + C} = e^C e^{kt} \] Letting \( A = e^C \), we have: \[ y = A e^{kt} \] Since \( y(0) = 1000 \): \[ 1000 = A e^{k \cdot 0} \implies A = 1000 \] So: \[ y = 1000 e^{kt} \] Now we use the second condition \( y(10) = 1500 \): \[ 1500 = 1000 e^{10k} \implies e^{10k} = \frac{1500}{1000} = 1.5 \] Taking the natural logarithm: \[ 10k = \ln(1.5) \implies k = \frac{\ln(1.5)}{10} \] Now we can find \( y(18) \): \[ y(18) = 1000 e^{18k} = 1000 e^{18 \cdot \frac{\ln(1.5)}{10}} = 1000 e^{\frac{18 \ln(1.5)}{10}} = 1000 (1.5)^{1.8} \] Calculating \( (1.5)^{1.8} \): \[ (1.5)^{1.8} \approx 2.148 \] Thus: \[ y(18) \approx 1000 \times 2.148 = 2148 \] Rounding to the nearest whole number, we find: \[ y \approx 2148 \] So, the final answer is: \[ y = 2148 \]