Let's analyze the circuit and find \( v(t) \) for \( t > 0 \): ### 1. **Identify Elements and Sources** - **Resistor**: \( 8\,\Omega \) - **Inductors**: \( 2\,\text{H} \), \( 4\,\text{H} \) - **Capacitor**: \( \frac{1}{2}\,\text{F} \) - **Voltage Source**: \( 5\sin(5t) \) - **Current Source**: \( 4\cos(3t+15^\circ) \) **A** (where \( v(t) \) is the voltage across this source) --- ### 2. **Write the Differential Equation** Let’s use the node-voltage method at the top node (where all elements meet). Let the voltage at the top node be \( v(t) \). #### **Branch Currents:** - Through the resistor: \( \frac{v(t) - 5\sin(5t)}{8} \) - Through the 2H inductor: \( \frac{1}{2} \int i_{2H} dt = v(t) \Rightarrow i_{2H} = \frac{1}{2} \frac{dv(t)}{dt} \) - Through the 4H inductor: \( i_{4H} = \frac{1}{4} \frac{dv(t)}{dt} \) - Through the capacitor: \( i_C = \frac{d}{dt}\left(C v(t)\right) = \frac{1}{2} \frac{dv(t)}{dt} \) But actually, looking at the configuration, let's write the KCL at the node where the current source is connected (assuming the bottom node is ground): \[ i_{8\Omega} + i_{2H} + i_{4H} + i_{C} = 4\cos(3t+15^\circ) \] But let's clarify each current: - Through \( 8\Omega \): \( i_{8\Omega} = \frac{v(t) - 5\sin(5t)}{8} \) - Through \( 2H \) inductor: \( i_{2H} = \frac{1}{2} \int_0^t v(\tau) d\tau \) - Through \( 4H \) inductor: \( i_{4H} = \frac{1}{4} \int_0^t v(\tau) d\tau \) - Through \( \frac{1}{2}F \) capacitor: \( i_C = \frac{1}{2} \frac{dv(t)}{dt} \) But since \( 2H \) and \( 4H \) are in parallel, let's find their equivalent inductance: \[ L_{eq} = \left( \frac{1}{2} + \frac{1}{4} \right)^{-1} = \frac{4}{3} \, H \] So, the current through the equivalent inductor: \[ i_L = \frac{1}{L_{eq}} \int_0^t v(\tau) d\tau = \frac{3}{4} \int_0^t v(\tau) d\tau \] So the node equation is: \[ \frac{v(t) - 5\sin(5t)}{8} + \frac{3}{4} \int_0^t v(\tau) d\tau + \frac{1}{2} \frac{dv(t)}{dt} = 4\cos(3t+15^\circ) \] --- ### 3. **Formulate the Differential Equation** Take the derivative of both sides to eliminate the integral: \[ \frac{1}{8} \frac{dv(t)}{dt} - \frac{5}{8} \cdot 5\cos(5t) + \frac{3}{4} v(t) + \frac{1}{2} \frac{d^2 v(t)}{dt^2} = -4 \cdot 3 \sin(3t + 15^\circ) \] \[ \frac{1}{8} \frac{dv(t)}{dt} + \frac{3}{4} v(t) + \frac{1}{2} \frac{d^2 v(t)}{dt^2} = 4\cos(3t+15^\circ) + \frac{25}{8}\cos(5t) \] --- ### 4. **Final Differential Equation** \[ \frac{1}{2} v''(t) + \frac{1}{8} v'(t) + \frac{3}{4} v(t) = 4\cos(3t+15^\circ) + \frac{25}{8}\cos(5t) \] Or, multiplying both sides by 8 to clear fractions: \[ 4 v''(t) + v'(t) + 6 v(t) = 32\cos(3t+15^\circ) + 25\cos(5t) \] --- ### 5. **Solution** The complete solution is: \[ v(t) = v_h(t) + v_p(t) \] Where: - \( v_h(t) \) is the homogeneous solution (solve the characteristic equation for \( 4r^2 + r + 6 = 0 \)) - \( v_p(t) \) is the particular solution (sum of two cosine functions at frequencies 3 and 5 rad/s, with unknown amplitudes and phases to be determined). --- #### **Homogeneous Solution:** \[ 4r^2 + r + 6 = 0 \] Solve for \( r \): \[ r = \frac{-1 \pm \sqrt{1 - 4 \cdot 4 \cdot 6}}{8} = \frac{-1 \pm \sqrt{1 - 96}}{8} = \frac{-1 \pm j\sqrt{95}}{8} \] So, \[ v_h(t) = e^{-t/8} \left[ A \cos\left(\frac{\sqrt{95}}{8} t\right) + B \sin\left(\frac{\sqrt{95}}{8} t\right) \right] \] --- #### **Particular Solution:** Try a solution of the form: \[ v_p(t) = C_1 \cos(3t + 15^\circ) + C_2 \sin(3t + 15^\circ) + D_1 \cos(5t) + D_2 \sin(5t) \] Plug into the equation and solve for \( C_1, C_2, D_1, D_2 \). --- ### **Summary of the Approach** 1. **Set up the differential equation as above.** 2. **Find the homogeneous solution.** 3. **Find the particular solution for the two frequencies.** 4. **Determine the initial conditions if provided, or solve for the constants using the initial behavior of the circuit.** If you need the explicit values for the coefficients in \( v_p(t) \) or the full step-by-step solution, let me know!