Certainly! Here is the content from the images, organized stepwise as requested: --- ### **Step 1** Given: \[ V(0^-) = 0 \text{ V} \] \[ i_L(0^-) = \frac{45}{40} = 1.125 \text{ A} \] At \( t > 0 \) [Switch is closed to B]: \[ \alpha = \frac{R}{2L} = \frac{10}{2 \times 4} = 1.25 \] \[ \omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{4 \times 0.25}} = 1 \] **Explanation:** Here, \[ \alpha > \omega_0 \] So, it is an **overdamped case**. \[ S_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} \] \[ = -1.25 \pm \sqrt{1.5625 - 1} \] \[ S_{1,2} = -0.5 \text{ and } -2 \] \[ V(t) = Ae^{-0.5t} + Ce^{-2t} \qquad \cdots (1) \] --- ### **Step 2** The current equation is given by: \[ i_C(0) = C \cdot \frac{dV(0)}{dt} = 1.125 \] Now solving: \[ \frac{dV(0)}{dt} = \frac{1.125}{C} \] \[ \frac{dV(0)}{dt} = \frac{1.125}{0.25} \] \[ \frac{dV(0)}{dt} = 4.5 \] **Explanation:** Differentiate the equation (1): \[ \frac{dV(t)}{dt} = -2Ae^{-0.5t} - 0.5Ce^{-2t} \] At \( t = 0 \): \[ \frac{dV(0)}{dt} = -2A - 0.5C \] From earlier: \[ \frac{dV(0)}{dt} = 4.5 \] Substitute \(C = -A\): \[ 4.5 = -2A - 0.5(-A) \] \[ 4.5 = -2A + 0.5A \] \[ A(-1.5) = 4.5 \] \[ A = -3 \] So, \[ A = -3 \implies C = +3 \] --- ### **Final Answer** The values are: \[ A = -3 \] \[ B = -0.5 \] \[ C = +3 \] \[ D = -2 \]