Let's solve the circuit step by step: --- ### **Step 1: Analyze the Circuit for \( t < 0 \) (Switch in Position \( a \))** - The circuit has been in position \( a \) for a long time. This means **steady-state** conditions. - For a long time: - **Inductor (1 H):** Acts as a short circuit (since \( \frac{di}{dt} = 0 \)). - **Capacitor (1 μF):** Acts as an open circuit (since \( i_C = C\frac{dv}{dt} = 0 \)). #### **Redrawing for \( t < 0 \):** - The 5 mA current source is in parallel with: - 4 kΩ - 100 Ω in series with (1 H shorted and 500 Ω in parallel, but 1 H is a short, so 500 Ω is also shorted). - The voltage across the capacitor \( v_0(0^-) \) is just the voltage across the open capacitor (across 2 kΩ, but since it's open, it's the node voltage). ##### **Find Initial Voltage Across Capacitor \( v_0(0^-) \):** - The capacitor is open, so no current through 2kΩ or 1μF. - All the current from the source splits between 4kΩ and 100Ω branches. \[ I_{total} = 5\,mA \] The voltage across the 4kΩ and 100Ω (which are in parallel): \[ V_{parallel} = I_{total} \left[ \frac{1}{\frac{1}{4000} + \frac{1}{100}} \right] \] Calculate the equivalent resistance: \[ \frac{1}{R_{eq}} = \frac{1}{4000} + \frac{1}{100} = \frac{1 + 40}{4000} = \frac{41}{4000} \] \[ R_{eq} = \frac{4000}{41} \approx 97.56 \,\Omega \] \[ V_{parallel} = 5 \times 10^{-3} \times 97.56 \approx 0.488\,V \] So, \[ v_0(0^-) = 0.488\,V \] --- ### **Step 2: At \( t = 0 \), Switch Moves to Position \( b \)** - Now, the left side is disconnected. - The right side circuit (for \( t \geq 0 \)): - 1 H inductor, 500 Ω, 2kΩ, and 1 μF capacitor. #### **Initial Conditions:** - **Voltage across capacitor can't change instantaneously:** \[ v_0(0^+) = v_0(0^-) = 0.488\,V \] - **Current through inductor can't change instantaneously:** - Find \( i_L(0^-) \): For \( t < 0 \): - Inductor is shorted. - The current through 100Ω is: \[ V_{parallel} = 0.488\,V \implies I_{100} = \frac{0.488}{100} = 4.88\,mA \] - All of this current flows through inductor (since it’s a short). \[ i_L(0^-) = 4.88\,mA \] \[ i_L(0^+) = i_L(0^-) = 4.88\,mA \] --- ### **Step 3: Formulate the Differential Equation for \( t \geq 0 \)** **Redraw the circuit for \( t \geq 0 \):** - The inductor (1H) in series with 500Ω, and that branch is in parallel with 2kΩ and 1μF capacitor. Let \( v_0(t) \) be the voltage across the capacitor. #### **Writing KCL at the Top Node:** \[ \text{Sum of currents leaving node:} \] 1. **Through \( 2k\Omega \):** \( \frac{v_0}{2000} \) 2. **Through Capacitor:** \( C \frac{dv_0}{dt} = 1 \mu F \frac{dv_0}{dt} \) 3. **Through Inductor Branch:** - Let current through inductor branch be \( i_L \). - KVL for inductor branch: \[ v_0 = L \frac{di_L}{dt} + 500 i_L \] \[ \Rightarrow \frac{di_L}{dt} = \frac{v_0 - 500 i_L}{1} \] - The current entering the top node from this branch is \( i_L \). \[ \frac{v_0}{2000} + 1 \times 10^{-6} \frac{dv_0}{dt} + i_L = 0 \] But we have two unknowns, \( v_0 \) and \( i_L \). #### **Express \( i_L \) in terms of \( v_0 \):** From above: \[ \frac{di_L}{dt} = v_0 - 500 i_L \] Or: \[ \frac{di_L}{dt} + 500 i_L = v_0 \] Now differentiate the KCL equation: \[ \frac{1}{2000} \frac{dv_0}{dt} + 1 \times 10^{-6} \frac{d^2 v_0}{dt^2} + \frac{di_L}{dt} = 0 \] But \( \frac{di_L}{dt} = v_0 - 500 i_L \): \[ \frac{1}{2000} \frac{dv_0}{dt} + 1 \times 10^{-6} \frac{d^2 v_0}{dt^2} + v_0 - 500 i_L = 0 \] But from the original KCL: \[ i_L = - \left( \frac{v_0}{2000} + 1 \times 10^{-6} \frac{dv_0}{dt} \right) \] Substitute \( i_L \): \[ \frac{1}{2000} \frac{dv_0}{dt} + 1 \times 10^{-6} \frac{d^2 v_0}{dt^2} + v_0 - 500 \left[ - \left( \frac{v_0}{2000} + 1 \times 10^{-6} \frac{dv_0}{dt} \right) \right] = 0 \] \[ \frac{1}{2000} \frac{dv_0}{dt} + 1 \times 10^{-6} \frac{d^2 v_0}{dt^2} + v_0 + \frac{500}{2000} v_0 + 500 \times 1 \times 10^{-6} \frac{dv_0}{dt} = 0 \] \[ \frac{1}{2000} \frac{dv_0}{dt} + 1 \times 10^{-6} \frac{d^2 v_0}{dt^2} + v_0 + 0.25 v_0 + 0.0005 \frac{dv_0}{dt} = 0 \] \[ 1 \times 10^{-6} \frac{d^2 v_0}{dt^2} + \left( \frac{1}{2000} + 0.0005 \right) \frac{dv_0}{dt} + (1 + 0.25) v_0 = 0 \] \[ 1 \times 10^{-6} \frac{d^2 v_0}{dt^2} + (0.0005 + 0.0005) \frac{dv_0}{dt} + 1.25 v_0 = 0 \] \[ 1 \times 10^{-6} \frac{d^2 v_0}{dt^2} + 0.001 \frac{dv_0}{dt} + 1.25 v_0 = 0 \] --- ### **Step 4: Solve the Differential Equation** Characteristic equation: \[ 1 \times 10^{-6} s^2 + 0.001 s + 1.25 = 0 \] \[ s^2 + 1000 s + 1.25 \times 10^6 = 0 \] Use quadratic formula: \[ s = \frac{-1000 \pm \sqrt{1000^2 - 4 \times 1 \times 1.25 \times 10^6}}{2} \] \[ = \frac{-1000 \pm \sqrt{1,000,000 - 5,000,000}}{2} \] \[ = \frac{-1000 \pm \sqrt{-4,000,000}}{2} \] \[ = \frac{-1000 \pm 2000j}{2} \] \[ = -500 \pm 1000j \] So the solution is: \[ v_0(t) = A e^{-500 t} \cos(1000 t) + B e^{-500 t} \sin(1000 t) \] --- ### **Step 5: Find Constants Using Initial Conditions** Recall initial conditions: - \( v_0(0) = 0.488\,V \) - \( \frac{dv_0}{dt}|_{t=0} \): Find using capacitor current and KCL. From KCL at \( t = 0 \): \[ \frac{v_0(0)}{2000} + 1 \times 10^{-6} \frac{dv_0}{dt}\bigg|_{t=0} + i_L(0^+) = 0 \] \[ \frac{0.488}{2000} + 1 \times 10^{-6} \frac{dv_0}{dt}\bigg|_{t=0} + 0.00488 = 0 \] \[ 0.000244 + 1 \times 10^{-6} \frac{dv_0}{dt}\bigg|_{t=0} + 0.00488 = 0 \] \[ 1 \times 10^{-6} \frac{dv_0}{dt}\bigg|_{t=0} = -0.005124 \] \[ \frac{dv_0}{dt}\bigg|_{t=0} = -5124\,\text{V/s} \] --- **Now substitute into the general solution:** \[ v_0(0) = A = 0.488 \] \[ \frac{dv_0}{dt} = -500A + 1000B \] \[ -5124 = -500 \times 0.488 + 1000B \] \[ -5124 = -244 + 1000B \] \[ 1000B = -5124 + 244 = -4880 \] \[ B = -4.88 \] --- ## **Final Answer** \[ \boxed{ v_0(t) = 0.488\,e^{-500 t} \cos(1000 t) - 4.88\,e^{-500 t} \sin(1000 t) \quad \text{(for } t \geq 0 \text{)} } \] --- If you need the answer in another form, or a plot, let me know!

Let's analyze the circuit step-by-step to find \( v_p(t) \): --- ### **Step 1: Steady-State Conditions for \( t < 0 \)** - The switch has been in position **a** for a long time. - **Inductor (1 H):** behaves as a short circuit in steady state. - **Capacitor (1 μF):** behaves as an open circuit in steady state. **Given:** - Current source: 5 mA - The circuit is connected such that the steady-state voltage across the capacitor can be found from the current source splitting. **Calculate the voltage across the capacitor at \( t = 0^- \):** - The 5 mA source feeds current into the parallel combination of: - 4 kΩ resistor - 100 Ω resistor in series with inductor (which is shorted in steady state) - The capacitor (open circuit in steady state) - The voltage across the parallel branch is: \[ V_{parallel} = I_{total} \times R_{eq} \] - Find \( R_{eq} \): \[ \frac{1}{R_{eq}} = \frac{1}{4000} + \frac{1}{100} = \frac{1 + 40}{4000} = \frac{41}{4000} \] \[ R_{eq} = \frac{4000}{41} \approx 97.56\, \Omega \] - Voltage across the capacitor: \[ v_0(0^-) = 5 \times 10^{-3} \times 97.56 \approx 0.488\,V \] **Initial capacitor voltage:** \[ v_0(0^-) = 0.488\, V \] --- ### **Step 2: Initial Conditions at \( t=0 \)** - Since the switch changes instantaneously, the capacitor voltage remains continuous: \[ v_0(0^+) = v_0(0^-) = 0.488\,V \] - The inductor current also remains the same: \[ i_L(0^-) = \frac{v_{parallel}}{R} = \frac{0.488}{100} \approx 4.88\, mA \] --- ### **Step 3: Formulating the Differential Equation for \( t \geq 0 \)** **Circuit after switching:** - The inductor (1 H) in series with 500 Ω resistor. - The capacitor (1 μF) is connected across the inductor-resistor series. - The 2 kΩ resistor is in parallel with the capacitor. **Define:** - \( v(t) \): voltage across the capacitor (and across the inductor-resistor network). **Applying KCL at the node:** \[ \frac{v(t)}{2000} + C \frac{dv(t)}{dt} + i_L(t) = 0 \] - \( i_L(t) \): current through the inductor, satisfying: \[ L \frac{di_L}{dt} + R_{L} i_L = v(t) \] with \( L=1\,H \), \( R_{L} = 500\,Ω \). --- ### **Step 4: Deriving the Differential Equation** Express \( i_L(t) \): \[ \frac{di_L}{dt} + 500 i_L = v(t) \] Solve this system: 1. From the KCL: \[ \frac{v(t)}{2000} + C \frac{dv(t)}{dt} + i_L(t) = 0 \] 2. Differentiate both sides: \[ \frac{1}{2000} \frac{dv(t)}{dt} + C \frac{d^2 v(t)}{dt^2} + \frac{di_L}{dt} = 0 \] 3. Substitute \( \frac{di_L}{dt} = v(t) - 500 i_L(t) \): \[ \frac{1}{2000} \frac{dv(t)}{dt} + C \frac{d^2 v(t)}{dt^2} + v(t) - 500 i_L(t) = 0 \] But from the KCL: \[ i_L(t) = - \left( \frac{v(t)}{2000} + C \frac{dv(t)}{dt} \right) \] Substitute into the previous: \[ \frac{1}{2000} \frac{dv(t)}{dt} + C \frac{d^2 v(t)}{dt^2} + v(t) + 500 \left( \frac{v(t)}{2000} + C \frac{dv(t)}{dt} \right) = 0 \] Simplify: \[ \frac{1}{2000} \frac{dv(t)}{dt} + C \frac{d^2 v(t)}{dt^2} + v(t) + \frac{500}{2000} v(t) + 500 C \frac{dv(t)}{dt} = 0 \] \[ C \frac{d^2 v(t)}{dt^2} + \left( \frac{1}{2000} + 500 C \right) \frac{dv(t)}{dt} + \left(1 + 0.25 \right) v(t) = 0 \] Given \( C = 1 \times 10^{-6} \): \[ C \frac{d^2 v}{dt^2} = 1 \times 10^{-6} \frac{d^2 v}{dt^2} \] \[ \frac{1}{2000} = 0.0005 \] \[ 500 C = 500 \times 10^{-6} = 0.0005 \] So: \[ 1 \times 10^{-6} \frac{d^2 v}{dt^2} + (0.0005 + 0.0005) \frac{dv}{dt} + 1.25 v = 0 \] \[ 1 \times 10^{-6} \frac{d^2 v}{dt^2} + 0.001 \frac{dv}{dt} + 1.25 v = 0 \] --- ### **Step 5: Solve the Differential Equation** Characteristic equation: \[ 1 \times 10^{-6} s^2 + 0.001 s + 1.25 = 0 \] Multiply through by \( 10^6 \): \[ s^2 + 1000 s + 1.25 \times 10^6 = 0 \] Discriminant: \[ \Delta = (1000)^2 - 4 \times 1.25 \times 10^6 = 1,000,000 - 5,000,000 = -4,000,000 \] Complex roots: \[ s = \frac{-1000 \pm j \sqrt{4,000,000}}{2} = -500 \pm j 1000 \] **Solution:** \[ v(t) = e^{-500 t} \left( K_1 \cos(1000 t) + K_2 \sin(1000 t) \right) \] --- ### **Step 6: Find Constants Using Initial Conditions** - \( v(0) = 0.488\,V \) - \( \frac{dv}{dt} \) at \( t=0 \): Differentiate: \[ \frac{dv}{dt} = e^{-500 t} \left[ -500 (K_1 \cos 1000 t + K_2 \sin 1000 t) + 1000 (-K_1 \sin 1000 t + K_2 \cos 1000 t) \right] \] At \( t=0 \): \[ \frac{dv}{dt}\bigg|_{t=0} = (K_1 \times 1000) - (K_2 \times 500) \] From initial voltage: \[ v(0) = K_1 = 0.488\,V \] From the current through the capacitor at \( t=0 \): \[ i_C(0) = C \frac{dv}{dt} \Rightarrow \frac{dv}{dt}\bigg|_{t=0} = \frac{i_{C}(0)}{C} \] - The initial capacitor current \( i_C(0) \): \[ i_C(0) = \frac{v(0)}{R_{parallel}} = \frac{0.488}{2000} \approx 0.000244\,A \] But more directly, from the circuit, the current through 2kΩ resistor is: \[ i_{res} = \frac{v(0)}{2000} = 0.000244\,A \] Since the capacitor current is: \[ i_C(0) = C \frac{dv}{dt} \] \[ \frac{dv}{dt}\bigg|_{t=0} = \frac{i_C(0)}{C} = \frac{0.000244}{1 \times 10^{-6}} = 244\,000\, \text{V/sec} \] Now, plug into the derivative expression: \[ 244,000 = 1000 \times 0.488 - 500 \times K_2 \] Solve for \( K_2 \): \[ 244,000 = 488 - 500 K_2 \] \[ 500 K_2 = 488 - 244,000 \approx -243,512 \] \[ K_2 \approx -487.02 \] --- ### **### **Final Expression:** \[ \boxed{ v(t) = 0.488 e^{-500 t} \cos(1000 t) - 487.02 e^{-500 t} \sin(1000 t) } \] --- ### **Summary:** - The voltage across the capacitor immediately after switching is approximately **0.488 V**. - The transient response involves exponentially damped sinusoidal oscillations with frequency \( \approx 1000\, \text{rad/sec} \) and damping factor \( 500\, \text{sec}^{-1} \). --- **Note:** The large magnitude of \( K_2 \) indicates a rapid initial change in voltage, consistent with the circuit's parameters. Let me know if you'd like a plot or further explanation!

Let's plot the voltage response \( v(t) \) based on the derived expression: \[ v(t) = 0.488\, e^{-500 t} \cos(1000 t) - 487.02\, e^{-500 t} \sin(1000 t) \] ### Plotting details: - Time range: from 0 to 5 milliseconds (\(0 \leq t \leq 0.005\,s\)) - Units: Volts for voltage, milliseconds for time ### Python code for plotting (if you want to generate the plot): ```python import numpy as np import matplotlib.pyplot as plt # Time array from 0 to 5 ms t = np.linspace(0, 0.005, 1000) # Compute v(t) v = (0.488 * np.exp(-500 * t) * np.cos(1000 * t) - 487.02 * np.exp(-500 * t) * np.sin(1000 * t)) # Plot plt.figure(figsize=(10, 6)) plt.plot(t * 1000, v) # convert t to milliseconds plt.title('Voltage across capacitor v(t) for 0 ≤ t ≤ 5 ms') plt.xlabel('Time (ms)') plt.ylabel('Voltage (V)') plt.grid(True) plt.show() ``` ### Explanation of the plot: - The voltage exhibits a damped oscillation starting at approximately 0.488 V. - The exponential decay with a time constant of \( \tau = 1/500 = 2\,ms \) causes the oscillations to diminish quickly. - The high frequency (~1000 rad/sec) results in rapid oscillations within the time window. If you'd like me to prepare the plot image directly or analyze further, please let me know!

Certainly! Here's the output of the code, which is a plot showing the voltage \( v(t) \) across the capacitor from \( t = 0 \) to \( t = 5\,ms \): --- **Voltage Response \( v(t) \) vs. Time (ms):**  *(Note: As an AI text-based model, I cannot generate actual images, but if you run the provided Python code in your environment, it will produce a plot similar to the following description.)* --- ### **Description of the plot:** - **X-axis:** Time in milliseconds (ms), ranging from 0 to 5 ms. - **Y-axis:** Voltage in volts (V). - **Curve:** Starts at approximately 0.488 V at \( t=0 \). - **Behavior:** - The voltage oscillates rapidly due to the high-frequency sinusoidal component (~1000 rad/sec). - The amplitude of oscillations diminishes exponentially over time with a time constant of about 2 ms. - After a few milliseconds, the voltage approaches zero as the transient effects die out. --- ### **Summary:** This plot visually demonstrates the transient damped oscillation response of the capacitor voltage after switching, consistent with the derived analytical solution. --- **If you'd like, I can help you generate the plot in your environment or interpret it further!**

Let's analyze the circuit step-by-step to find the voltage \( v_p(t) \): --- ### **Step 1: Steady-State Conditions for \( t < 0 \)** - The switch has been in position **a** for a long time. - **Inductor (1 H):** behaves as a short circuit in steady state. - **Capacitor (1 μF):** behaves as an open circuit in steady state. **Given:** - Current source: 5 mA - The circuit is connected such that the steady-state voltage across the capacitor can be found from the current source splitting. **Calculate the voltage across the capacitor at \( t = 0^- \):** - The 5 mA source feeds current into the parallel combination of: - 4 kΩ resistor - 100 Ω resistor in series with inductor (which is shorted in steady state) - The capacitor (open circuit in steady state) - The voltage across the parallel branch is: \[ V_{parallel} = I_{total} \times R_{eq} \] - Find \( R_{eq} \): \[ \frac{1}{R_{eq}} = \frac{1}{4000} + \frac{1}{100} = \frac{1 + 40}{4000} = \frac{41}{4000} \] \[ R_{eq} = \frac{4000}{41} \approx 97.56\, \Omega \] - Voltage across the capacitor: \[ v_0(0^-) = 5 \times 10^{-3} \times 97.56 \approx 0.488\,V \] **Initial capacitor voltage:** \[ v_0(0^-) = 0.488\, V \] --- ### **Step 2: Initial Conditions at \( t=0 \)** - Since the switch changes instantaneously, the capacitor voltage remains continuous: \[ v_0(0^+) = v_0(0^-) = 0.488\,V \] - The inductor current also remains the same: \[ i_L(0^-) = \frac{v_{parallel}}{R} = \frac{0.488}{100} \approx 4.88\, mA \] --- ### **Step 3: Formulating the Differential Equation for \( t \geq 0 \)** **Circuit after switching:** - The inductor (1 H) in series with 500 Ω resistor. - The capacitor (1 μF) is connected across the inductor-resistor series. - The 2 kΩ resistor is in parallel with the capacitor. **Define:** - \( v(t) \): voltage across the capacitor (and across the inductor-resistor network). **Applying KCL at the node:** \[ \frac{v(t)}{2000} + C \frac{dv(t)}{dt} + i_L(t) = 0 \] - \( i_L(t) \): current through the inductor, satisfying: \[ L \frac{di_L}{dt} + R_{L} i_L = v(t) \] with \( L=1\,H \), \( R_{L} = 500\,Ω \). --- ### **Step 4: Deriving the Differential Equation** Express \( i_L(t) \): \[ \frac{di_L}{dt} + 500 i_L = v(t) \] Solve this system: 1. From the KCL: \[ \frac{v(t)}{2000} + C \frac{dv(t)}{dt} + i_L(t) = 0 \] 2. Differentiate both sides: \[ \frac{1}{2000} \frac{dv(t)}{dt} + C \frac{d^2 v(t)}{dt^2} + \frac{di_L}{dt} = 0 \] 3. Substitute \( \frac{di_L}{dt} = v(t) - 500 i_L(t) \): \[ \frac{1}{2000} \frac{dv(t)}{dt} + C \frac{d^2 v(t)}{dt^2} + v(t) - 500 i_L(t) = 0 \] But from the KCL: \[ i_L(t) = - \left( \frac{v(t)}{2000} + C \frac{dv(t)}{dt} \right) \] Substitute into the previous: \[ \frac{1}{2000} \frac{dv(t)}{dt} + C \frac{d^2 v(t)}{dt^2} + v(t) + 500 \left( \frac{v(t)}{2000} + C \frac{dv(t)}{dt} \right) = 0 \] Simplify: \[ C \frac{d^2 v}{dt^2} + \left( \frac{1}{2000} + 500 C \right) \frac{dv}{dt} + \left(1 + 0.25 \right) v(t) = 0 \] Given \( C = 1 \times 10^{-6} \): \[ C \frac{d^2 v}{dt^2} = 1 \times 10^{-6} \frac{d^2 v}{dt^2} \] \[ \frac{1}{2000} = 0.0005 \] \[ 500 C = 500 \times 10^{-6} = 0.0005 \] So: \[ 1 \times 10^{-6} \frac{d^2 v}{dt^2} + (0.0005 + 0.0005) \frac{dv}{dt} + 1.25 v(t) = 0 \] \[ 1 \times 10^{-6} \frac{d^2 v}{dt^2} + 0.001 \frac{dv}{dt} + 1.25 v(t) = 0 \] --- ### **Step 5: Solve the Differential Equation** Characteristic equation: \[ 1 \times 10^{-6} s^2 + 0.001 s + 1.25 = 0 \] Multiply through by \( 10^6 \): \[ s^2 + 1000 s + 1.25 \times 10^6 = 0 \] Discriminant: \[ \Delta = (1000)^2 - 4 \times 1.25 \times 10^6 = 1,000,000 - 5,000,000 = -4,000,000 \] Complex roots: \[ s = \frac{-1000 \pm j \sqrt{4,000,000}}{2} = -500 \pm j 1000 \] **Solution:** \[ v(t) = e^{-500 t} \left( K_1 \cos(1000 t) + K_2 \sin(1000 t) \right) \] --- ### **Step 6: Find Constants Using Initial Conditions** - \( v(0) = 0.488\,V \) - \( \frac{dv}{dt} \) at \( t=0 \): Differentiate: \[ \frac{dv}{dt} = e^{-500 t} \left[ -500 (K_1 \cos 1000 t + K_2 \sin 1000 t) + 1000 (-K_1 \sin 1000 t + K_2 \cos 1000 t) \right] \] At \( t=0 \): \[ \frac{dv}{dt}\bigg|_{t=0} = (K_1 \times 1000) - (K_2 \times 500) \] From initial voltage: \[ v(0) = K_1 = 0.488\,V \] From the initial capacitor current: \[ i_C(0) = C \frac{dv}{dt} \Rightarrow \frac{dv}{dt}\bigg|_{t=0} = \frac{i_{C}(0)}{C} \] - The initial capacitor current \( i_C(0) \): \[ i_C(0) = \frac{v(0)}{R_{parallel}} = \frac{0.488}{2000} \approx 0.000244\,A \] Thus: \[ \frac{dv}{dt}\bigg|_{t=0} = \frac{0.000244}{1 \times 10^{-6}} = 244\,000\, \text{V/sec} \] Now plug into the derivative expression: \[ 244,000 = 1000 \times 0.488 - 500 K_2 \] \[ 244,000 = 488 - 500 K_2 \] Solve for \( K_2 \): \[ 500 K_2 = 488 - 244,000 \approx -243,512 \] \[ K_2 \approx -487.02 \] --- ### **Final Expression:** \[ \boxed{ v(t) = 0.488\, e^{-500 t} \cos(1000 t) - 487.02\, e^{-500 t} \sin(1000 t) } \] --- ### **Summary:** - The voltage across the capacitor immediately after switching is approximately **0.488 V**. - The transient response involves exponentially damped sinusoidal oscillations with frequency \( \approx 1000\, \text{rad/sec} \) and damping factor \( 500\, \text{sec}^{-1} \). --- **Note:** The large magnitude of \( K_2 \) indicates a rapid initial change in voltage, consistent with the circuit's parameters. Let me know if you'd like a plot or further explanation!