Let's solve the initial value problem using **Euler's Method** step-by-step. ### **Given:** \[ \frac{dy}{dx} - 5y = x \] \[ y(0) = 1 \] Estimate \(y(x)\) on the interval \([0, 1]\) with **5 steps**. --- ### **Step 1: Rewrite as Explicit ODE** \[ \frac{dy}{dx} = x + 5y \] Let \( f(x, y) = x + 5y \). --- ### **Step 2: Determine Step Size** The interval is \([0, 1]\) with 5 steps: \[ h = \frac{1 - 0}{5} = 0.2 \] --- ### **Step 3: Euler's Method Formula** \[ y_{n+1} = y_n + h \cdot f(x_n, y_n) \] --- ### **Step 4: Initialize** - \( x_0 = 0 \), \( y_0 = 1 \) - \( h = 0.2 \) --- ### **Step 5: Apply Euler's Method Iteratively** Let’s compute each step: #### **Step 0:** \[ x_0 = 0, \quad y_0 = 1 \] #### **Step 1:** \[ y_1 = y_0 + h \cdot f(x_0, y_0) = 1 + 0.2 \cdot (0 + 5 \cdot 1) = 1 + 0.2 \cdot 5 = 1 + 1 = 2 \] \[ x_1 = x_0 + h = 0.2 \] #### **Step 2:** \[ y_2 = y_1 + h \cdot f(x_1, y_1) = 2 + 0.2 \cdot (0.2 + 5 \cdot 2) = 2 + 0.2 \cdot (0.2 + 10) = 2 + 0.2 \cdot 10.2 = 2 + 2.04 = 4.04 \] \[ x_2 = 0.4 \] #### **Step 3:** \[ y_3 = y_2 + h \cdot f(x_2, y_2) = 4.04 + 0.2 \cdot (0.4 + 5 \cdot 4.04) = 4.04 + 0.2 \cdot (0.4 + 20.2) = 4.04 + 0.2 \cdot 20.6 = 4.04 + 4.12 = 8.16 \] \[ x_3 = 0.6 \] #### **Step 4:** \[ y_4 = y_3 + h \cdot f(x_3, y_3) = 8.16 + 0.2 \cdot (0.6 + 5 \cdot 8.16) = 8.16 + 0.2 \cdot (0.6 + 40.8) = 8.16 + 0.2 \cdot 41.4 = 8.16 + 8.28 = 16.44 \] \[ x_4 = 0.8 \] #### **Step 5:** \[ y_5 = y_4 + h \cdot f(x_4, y_4) = 16.44 + 0.2 \cdot (0.8 + 5 \cdot 16.44) = 16.44 + 0.2 \cdot (0.8 + 82.2) = 16.44 + 0.2 \cdot 83.0 = 16.44 + 16.6 = 33.04 \] \[ x_5 = 1.0 \] --- ### **Step 6: Final Results Table** | \( n \) | \( x_n \) | \( y_n \) | |---|-------|--------| | 0 | 0.0 | 1.00 | | 1 | 0.2 | 2.00 | | 2 | 0.4 | 4.04 | | 3 | 0.6 | 8.16 | | 4 | 0.8 | 16.44 | | 5 | 1.0 | 33.04 | --- ## **Final Answer:** Using Euler's Method with 5 steps (\(h = 0.2\)), the estimated values are: \[ y(0) = 1, \quad y(0.2) \approx 2, \quad y(0.4) \approx 4.04, \quad y(0.6) \approx 8.16, \quad y(0.8) \approx 16.44, \quad y(1.0) \approx 33.04 \]