Let's solve each question step by step: --- ### **1. RC Circuit Differential Equation** Given: - \( R = 0.5 \, \Omega \) - \( C = 2 \, \text{F} \) - \( E(t) = 50 - t \) - Initial condition: \( q(0) = 0 \) The differential equation: \[ R \frac{dq}{dt} + \frac{1}{C}q = E(t) \] Substitute the values: \[ 0.5 \frac{dq}{dt} + \frac{1}{2}q = 50 - t \] \[ \frac{dq}{dt} + q = 100 - 2t \] This is a linear first-order ODE. Let's solve it. #### **Step 1: Find the Integrating Factor** The integrating factor (I.F.) is: \[ \mu(t) = e^{\int 1 dt} = e^{t} \] #### **Step 2: Multiply the Equation by I.F.** \[ e^{t}\frac{dq}{dt} + e^{t}q = (100 - 2t)e^{t} \] \[ \frac{d}{dt}(e^{t}q) = (100 - 2t)e^{t} \] #### **Step 3: Integrate Both Sides** \[ e^{t}q = \int (100 - 2t)e^{t}dt \] Break the integral: \[ \int(100 - 2t)e^{t}dt = 100\int e^{t}dt - 2\int t e^{t}dt \] Recall: \(\int t e^{t}dt = (t-1)e^{t} + C\) So: \[ 100\int e^{t}dt = 100e^{t} \] \[ -2\int t e^{t}dt = -2[(t-1)e^{t}] = -2t e^{t} + 2e^{t} \] Sum: \[ 100e^{t} - 2t e^{t} + 2e^{t} = (102 - 2t)e^{t} \] So, \[ e^{t}q = (102 - 2t)e^{t} + C \] \[ q(t) = 102 - 2t + Ce^{-t} \] #### **Step 4: Apply Initial Condition** \( q(0) = 0 \): \[ 0 = 102 - 0 + C \implies C = -102 \] #### **Final Answer:** \[ \boxed{q(t) = 102 - 2t - 102e^{-t}} \] --- ### **2. Newton's Law of Cooling** Given: - Room temp: \( 70^\circ F \) - Outside temp: \( 50^\circ F \) (\( T_a \)) - \( T(0) = 70 \) - \( T(0.5) = 65 \) (\( t = 0.5 \) min) - Newton's Law: \(\frac{dT}{dt} = k(T_a - T)\) #### **(a) Initial Value Problem** \[ \frac{dT}{dt} = k(50 - T), \quad T(0) = 70 \] #### **(b) Explicit Solution** This is a separable equation: \[ \frac{dT}{dt} + kT = 50k \] Integrating factor: \[ \mu(t) = e^{kt} \] Multiply both sides: \[ e^{kt}\frac{dT}{dt} + ke^{kt}T = 50k e^{kt} \] \[ \frac{d}{dt}(e^{kt}T) = 50k e^{kt} \] Integrate: \[ e^{kt}T = 50 e^{kt} + C \] \[ T = 50 + Ce^{-kt} \] Initial condition \( T(0) = 70 \): \[ 70 = 50 + C \implies C = 20 \] So, \[ T(t) = 50 + 20 e^{-kt} \] Use \( T(0.5) = 65 \) to solve for \( k \): \[ 65 = 50 + 20 e^{-0.5k} \] \[ 15 = 20 e^{-0.5k} \] \[ \frac{15}{20} = e^{-0.5k} \] \[ 0.75 = e^{-0.5k} \] \[ \ln(0.75) = -0.5k \] \[ k = -2 \ln(0.75) \] \[ k \approx -2(-0.2877) \approx 0.5754 \] #### **Final Explicit Solution** \[ \boxed{T(t) = 50 + 20 e^{-0.5754t}} \] --- #### **(c) Thermometer Reading at \( t = 1 \) min** \[ T(1) = 50 + 20 e^{-0.5754 \times 1} \] \[ e^{-0.5754} \approx 0.562 \] \[ T(1) = 50 + 20 \times 0.562 = 50 + 11.24 = 61.24^\circ F \] \[ \boxed{T(1) \approx 61.2^\circ F} \] --- If you need more detailed steps or explanations for any part, let me know!

1. \( \boxed{q(t) = 102 - 2t - 102 e^{-t}} \) 2. (a) \( \frac{dT}{dt} = k(50 - T), \quad T(0) = 70 \) (b) \( T(t) = 50 + 20 e^{-0.5754 t} \) (c) \( T(1) \approx 61.2^\circ F \)