In Exercises \( 11-14 \), find the solution of the differential equation dydt \( =k y, k \) a constant, that satisfies the given conditions. \( \begin{array}{ll}\text { 11. } k=1.5, y(0)=100 & \text { 12. } k=-0.5, y(0)=200 \\ \text { 13. } y(0)=50, y(5)=100 & \text { 14. } y(0)=60, y(10)=30\end{array} \)

To solve the differential equation \(\frac{dy}{dt} = ky\), we start with the general solution, which is given by: \[ y(t) = Ce^{kt} \] where \(C\) is the constant determined by initial conditions. Now, let’s apply this to your specific exercises. 1. For \(k = 1.5\) and \(y(0) = 100\): \[ y(0) = C e^{1.5 \cdot 0} \implies 100 = C \implies C = 100. \] Therefore, the solution is: \[ y(t) = 100 e^{1.5t}. \] 2. For \(k = -0.5\) and \(y(0) = 200\): \[ y(0) = C e^{-0.5 \cdot 0} \implies 200 = C \implies C = 200. \] Hence, the solution is: \[ y(t) = 200 e^{-0.5t}. \] 3. For \(y(0) = 50\) and \(y(5) = 100\): Start with \(y(0) = C\): \[ y(0) = C = 50. \] Next, set up the second condition: \[ 100 = 50 e^{kt} \implies e^{5k} = 2 \implies 5k = \ln(2) \implies k = \frac{\ln(2)}{5}. \] Thus, the solution is: \[ y(t) = 50 e^{\frac{\ln(2)}{5}t}. \] 4. For \(y(0) = 60\) and \(y(10) = 30\): Again, start with \(y(0) = C\): \[ C = 60. \] Setting up the second condition: \[ 30 = 60 e^{kt} \implies e^{10k} = \frac{1}{2} \implies 10k = \ln\left(\frac{1}{2}\right) \implies k = -\frac{\ln(2)}{10}. \] Thus, the solution is: \[ y(t) = 60 e^{-\frac{\ln(2)}{10}t}. \]