The rate of change of \( y \) is directly proportional to \( y \). That is: \[ \frac{d y}{d t}=k y \] If \( y=1 \) when \( t=0 \) and \( y=2 \) when \( t=2 \), then find \( y \) when \( t=3 \). Do not round the value of but round your final answer to the nearest thousandth. \[ y=[\text { ? }] \]

The differential equation \(\frac{dy}{dt} = k y\) indicates that \(y\) grows exponentially. To solve this, we first separate variables and integrate: \[ \frac{1}{y} dy = k dt \] Integrating both sides gives: \[ \ln |y| = kt + C \] Exponentiating results in: \[ y = e^{kt + C} = e^{C} e^{kt} \] Letting \(C' = e^{C}\), we rewrite this as: \[ y = C' e^{kt} \] Next, we can use the initial condition \(y(0) = 1\) to find \(C'\): \[ 1 = C' e^{k \cdot 0} \implies C' = 1 \] Thus, we have: \[ y = e^{kt} \] Now, we can find \(k\) using the information \(y(2) = 2\): \[ 2 = e^{2k} \implies e^{2k} = 2 \] Taking the natural logarithm: \[ 2k = \ln(2) \implies k = \frac{\ln(2)}{2} \] Substituting back into our equation for \(y\): \[ y = e^{\frac{\ln(2)}{2} t} \] This can also be expressed as: \[ y = 2^{t/2} \] Now we want to find \(y\) when \(t = 3\): \[ y(3) = 2^{3/2} = 2^{1.5} = \sqrt{2^3} = \sqrt{8} = 2\sqrt{2} \approx 2.828 \] Rounding to the nearest thousandth, we have: \[ y \approx 2.828 \] So the final answer is: \[ y \approx [2.828] \]