An electric circuit contains a 1-H inductor, a \( 4-\Omega \) resistor, and a voltage source of cos \( t \). differential equation relating the current \( i \) and the time \( t \) is \( \mathrm{di} / \mathrm{dt}+4 \mathrm{i}=\cos \mathrm{t} \). Find \( i \) after method with \( \Delta \mathrm{t}=0.1 \mathrm{~s} \) if the initial current is zero. Solve the equation exactly and comp Use Euler's method to find \( i \) after 0.5 s . approx \( =0.2192 \) A (Round the final answer to four decimal places as needed. Round all intermediate values decimal places as needed.) Solve the equation exactly to find \( i \) after 0.5 s . iexact \( =\square \) A (Round to four decimal places as needed.)

To solve the problem, we will follow these steps: 1. **Solve the differential equation exactly** to find the current \( i(t) \). 2. **Use Euler's method** to approximate the current \( i \) after \( 0.5 \) seconds. 3. **Compare the results** from both methods. ### Step 1: Solve the Differential Equation Exactly The given differential equation is: \[ \frac{di}{dt} + 4i = \cos(t) \] This is a first-order linear ordinary differential equation. We can solve it using an integrating factor. 1. **Find the integrating factor** \( \mu(t) \): \[ \mu(t) = e^{\int 4 \, dt} = e^{4t} \] 2. **Multiply the entire equation by the integrating factor**: \[ e^{4t} \frac{di}{dt} + 4e^{4t} i = e^{4t} \cos(t) \] 3. **Rewrite the left side** as a derivative: \[ \frac{d}{dt}(e^{4t} i) = e^{4t} \cos(t) \] 4. **Integrate both sides**: \[ \int \frac{d}{dt}(e^{4t} i) \, dt = \int e^{4t} \cos(t) \, dt \] The right side can be solved using integration by parts or a known formula. The integral of \( e^{at} \cos(bt) \) is given by: \[ \int e^{at} \cos(bt) \, dt = \frac{e^{at}(a \cos(bt) + b \sin(bt))}{a^2 + b^2} \] In our case, \( a = 4 \) and \( b = 1 \): \[ \int e^{4t} \cos(t) \, dt = \frac{e^{4t}(4 \cos(t) + \sin(t))}{17} + C \] 5. **Substituting back**: \[ e^{4t} i = \frac{e^{4t}(4 \cos(t) + \sin(t))}{17} + C \] 6. **Solve for \( i(t) \)**: \[ i(t) = \frac{4 \cos(t) + \sin(t)}{17} + Ce^{-4t} \] 7. **Apply the initial condition** \( i(0) = 0 \): \[ 0 = \frac{4 \cos(0) + \sin(0)}{17} + C \implies 0 = \frac{4}{17} + C \implies C = -\frac{4}{17} \] Thus, the exact solution is: \[ i(t) = \frac{4 \cos(t) + \sin(t)}{17} - \frac{4}{17} e^{-4t} \] ### Step 2: Find \( i(0.5) \) Exactly Now we will calculate \( i(0.5) \): \[ i(0.5) = \frac{4 \cos(0.5) + \sin(0.5)}{17} - \frac{4}{17} e^{-2} \] Calculating the values: - \( \cos(0.5) \approx 0.8776 \) - \( \sin(0.5) \approx 0.4794 \) - \( e^{-2} \approx 0.1353 \) Now substituting these values: \[ i(0.5) = \frac{4(0.8776) + 0.4794}{17} - \frac{4(0.1353)}{17} \] Calculating the numerator: \[ 4(0.8776) + 0.4794 = 3.5104 + 0.4794 = 3.9898 \] Now substituting back: \[ i(0.5) = \frac{3.9898}{17} - \frac{0.5412}{17} = \frac{3.9898 - 0.5412}{17} = \frac{3.4486}{17} \approx 0.2028 \] ### Step 3: Use Euler's Method Using Euler's method with \( \Delta t = 0.1 \) for \( t = 0 \) to \( t = 0.5 \): 1. **Initial condition**: \( i(0) = 0 \) 2. **Update rule**: \( i_{n+1} = i_n + \Delta t \cdot f(t_n, i_n) \) where \( f(t, i) = \cos(t) - 4i \) Now we will calculate \( i \) at each step: - For \( t_0 = 0 \), \( i_0 = 0 \): \[ f(0, 0) = \cos(0) - 4(0) = 1 \] \[ i_1 = 0 + 0.1 \cdot 1 = 0.1 \] - For \( t_1 = 0.1 \), \( i_1 = 0.1 \): \[ f(0.1, 0.1) = \cos(0.1) - 4(0.1) \approx 0.995 - 0.4 = 0.595 \] \[ i_2 = 0.1 + 0.1 \cdot 0.595 = 0.1595 \] - For \( t_2 = 0.2 \), \( i_2 = 0.1595 \): \[ f(0.2, 0.1595) = \cos(0.2) - 4(0.1595) \approx 0.980 - 0.638 = 0.342 \] \[ i_3 = 0.1595 + 0.1 \cdot 0.342 = 0.1938 \] - For \( t_3 = 0.3 \), \( i_3 = 0.1938 \): \[ f(0.3, 0.1938) = \cos(0.3) - 4(0.1938) \approx 0.955 - 0.7752 = 0.1798 \] \[ i_4 = 0.1938 + 0.1 \cdot 0.1798 = 0.2118 \] - For \( t_4 = 0.4 \), \( i_4 = 0.211