1. **Problem Statement:** Find the inductor current $l(t)$ and the voltage $v_x(t)$ across the 1 $\Omega$ resistor in the given circuit, with initial condition $l(0) = 5$ A. 2. **Given Elements and Variables:** - Inductor $L = 2$ H - Resistors: $4 \Omega$, $1 \Omega$, and $4 \Omega$ - Dependent current source: $2i_x$ - Initial current through inductor: $l(0) = 5$ A - Voltage $v_x$ is across the $1 \Omega$ resistor in series with the dependent current source. 3. **Formulas and Rules:** - Voltage across an inductor: $$v_L = L \frac{di}{dt}$$ - Ohm's law for resistors: $$v = iR$$ - For the dependent current source, $i_x$ is the current through the $1 \Omega$ resistor. 4. **Step-by-step Solution:** 1. Let $l(t)$ be the current through the inductor. 2. The voltage across the inductor is $$v_L = 2 \frac{dl}{dt}$$. 3. The current $i_x$ through the $1 \Omega$ resistor is the same as the current through the dependent source branch, which is $i_x = l(t)$ because the dependent source is $2i_x$ and the current in the loop relates to $l(t)$. 4. The voltage across the $1 \Omega$ resistor is $$v_x = 1 \times i_x = l(t)$$. 5. Applying Kirchhoff's Voltage Law (KVL) around the loop: $$4i + v_x + 4i + v_L = 0$$ where $i$ is the loop current, which equals $l(t)$. 6. Substitute $v_x = l(t)$ and $v_L = 2 \frac{dl}{dt}$: $$4l(t) + l(t) + 4l(t) + 2 \frac{dl}{dt} = 0$$ 7. Simplify: $$9l(t) + 2 \frac{dl}{dt} = 0$$ 8. Rearrange the differential equation: $$2 \frac{dl}{dt} = -9l(t)$$ 9. Divide both sides by 2: $$\frac{dl}{dt} = -\frac{9}{2} l(t)$$ 10. This is a first-order linear differential equation with solution: $$l(t) = l(0) e^{-\frac{9}{2} t} = 5 e^{-4.5 t}$$ 11. Voltage across the $1 \Omega$ resistor is: $$v_x(t) = l(t) = 5 e^{-4.5 t}$$ **Final answers:** $$l(t) = 5 e^{-4.5 t} \quad \text{A}$$ $$v_x(t) = 5 e^{-4.5 t} \quad \text{V}$$ Note: The user-provided answer $5e^{-4t}$ and $-20e^{-4t}$ differs, possibly due to different assumptions or circuit details. Based on the given data and standard analysis, the above solution follows.