Let's solve the given problem step by step: --- ## **Given:** \[ \frac{dy}{dx} = x + y, \quad y(0) = 1, \quad T = 0.01 \] We are to: 1. Use **Euler’s Method** and **Heun’s Method** to estimate \( y(0.1) \). 2. Solve **analytically** and compare the results. --- ### **Step 1: Analytical Solution** This is a first-order linear ODE: \[ \frac{dy}{dx} - y = x \] The integrating factor (IF) is: \[ IF = e^{-\int 1 dx} = e^{-x} \] Multiply both sides by IF: \[ e^{-x}\frac{dy}{dx} - e^{-x}y = x e^{-x} \] \[ \frac{d}{dx}(y e^{-x}) = x e^{-x} \] Integrate both sides: \[ y e^{-x} = \int x e^{-x} dx + C \] Use integration by parts for \( \int x e^{-x} dx \): Let \( u = x \), \( dv = e^{-x} dx \), so \( du = dx \), \( v = -e^{-x} \): \[ \int x e^{-x} dx = -x e^{-x} + \int e^{-x} dx = -x e^{-x} - e^{-x} + C \] So, \[ y e^{-x} = -x e^{-x} - e^{-x} + C \] \[ y = -x - 1 + C e^{x} \] Use initial condition \( y(0) = 1 \): \[ 1 = -0 - 1 + C \implies C = 2 \] **Final Analytical Solution:** \[ \boxed{y(x) = -x - 1 + 2 e^{x}} \] Calculate at \( x = 0.1 \): \[ y(0.1) = -0.1 - 1 + 2e^{0.1} \] \[ e^{0.1} \approx 1.10517 \] \[ y(0.1) \approx -1.1 + 2 \times 1.10517 = -1.1 + 2.21034 = 1.11034 \] --- ### **Step 2: Euler’s Method** Euler’s formula: \( y_{n+1} = y_n + h f(x_n, y_n) \) Given: \( f(x, y) = x + y \), \( h = 0.01 \), \( y_0 = 1 \), \( x_0 = 0 \) We want \( y \) at \( x = 0.1 \) (i.e., after 10 steps). Let’s compute step by step: #### **Step-by-step Calculation Table:** | n | \(x_n\) | \(y_n\) | \(f(x_n, y_n)\) | \(y_{n+1}\) | |---|--------|---------|----------------|-----------| | 0 | 0.00 | 1.00000 | 1.00000 | 1.01000 | | 1 | 0.01 | 1.01000 | 1.02000 | 1.02020 | | 2 | 0.02 | 1.02020 | 1.04020 | 1.03060 | | 3 | 0.03 | 1.03060 | 1.06060 | 1.04121 | | 4 | 0.04 | 1.04121 | 1.08121 | 1.05202 | | 5 | 0.05 | 1.05202 | 1.10202 | 1.06304 | | 6 | 0.06 | 1.06304 | 1.12304 | 1.07427 | | 7 | 0.07 | 1.07427 | 1.14427 | 1.08571 | | 8 | 0.08 | 1.08571 | 1.16571 | 1.09737 | | 9 | 0.09 | 1.09737 | 1.18737 | 1.10924 | So, after 10 steps (\( x = 0.1 \)), **Euler’s Method:** \[ \boxed{y_{Euler}(0.1) \approx 1.10924} \] --- ### **Step 3: Heun’s Method** Heun’s method (Improved Euler): \[ y_{n+1} = y_n + \frac{h}{2} [f(x_n, y_n) + f(x_{n+1}, y^*_n)] \] where \( y^*_n = y_n + h f(x_n, y_n) \) is the Euler predictor. Let's do the first step explicitly: #### **First Step:** - \( x_0 = 0, y_0 = 1 \) - \( f(x_0, y_0) = 1 \) - Predictor: \( y^*_0 = 1 + 0.01 \times 1 = 1.01 \) - \( x_1 = 0.01 \), \( f(x_1, y^*_0) = 0.01 + 1.01 = 1.02 \) - Corrector: \( y_1 = 1 + \frac{0.01}{2}(1 + 1.02) = 1 + 0.0051 = 1.0051 \) #### **Repeat for next steps:** Continue this process up to \( x = 0.1 \) (10 steps). #### **Step-by-step Calculation Table:** | n | \(x_n\) | \(y_n\) | \(f(x_n, y_n)\) | \(y^*_n\) | \(f(x_{n+1}, y^*_n)\) | \(y_{n+1}\) | |---|--------|---------|----------------|---------|----------------------|-----------| | 0 | 0.00 | 1.00000 | 1.00000 | 1.01000 | 1.02000 | 1.00510 | | 1 | 0.01 | 1.00510 | 1.01510 | 1.01525 | 1.03525 | 1.02028 | | 2 | 0.02 | 1.02028 | 1.04028 | 1.03068 | 1.05068 | 1.03569 | | 3 | 0.03 | 1.03569 | 1.06569 | 1.04635 | 1.06635 | 1.05132 | | 4 | 0.04 | 1.05132 | 1.09132 | 1.06223 | 1.08223 | 1.06717 | | 5 | 0.05 | 1.06717 | 1.11717 | 1.07834 | 1.09834 | 1.08324 | | 6 | 0.06 | 1.08324 | 1.14324 | 1.09467 | 1.11467 | 1.09953 | | 7 | 0.07 | 1.09953 | 1.16953 | 1.11123 | 1.13123 | 1.11605 | | 8 | 0.08 | 1.11605 | 1.19605 | 1.12794 | 1.14794 | 1.13280 | | 9 | 0.09 | 1.13280 | 1.22280 | 1.14483 | 1.16483 | 1.14977 | So, after 10 steps (\( x = 0.1 \)), **Heun’s Method:** \[ \boxed{y_{Heun}(0.1) \approx 1.14977} \] --- ## **Step 4: Comparison Table** | Method | \(y(0.1)\) | |-----------------|---------------| | Analytical | 1.11034 | | Euler | 1.10924 | | Heun | 1.14977 | --- ### **Summary:** - **Euler’s method** gives \( y(0.1) \approx 1.10924 \) (very close to exact). - **Heun’s method** gives \( y(0.1) \approx 1.14977 \) (slightly overestimates). - **Exact solution** is \( y(0.1) \approx 1.11034 \). Both methods are close for this small step size, with Euler slightly underestimating and Heun slightly overestimating the result. --- **If you need the full step-by-step calculations for each iteration, I can provide those in detail as well!**

Certainly! Below is a detailed step-by-step solution, including all calculations for each step, for applying Euler’s and Heun’s methods to the differential equation, as well as the analytical solution and comparison. --- ## **Problem Recap:** Solve the differential equation: \[ \frac{dy}{dx} = x + y, \quad y(0) = 1 \] using: - Analytical method - Euler’s method with step size \( h = 0.01 \) - Heun’s method with step size \( h = 0.01 \) Calculate \( y \) at \( x = 0.1 \). --- ## **1. Analytical Solution** This is a linear differential equation: \[ \frac{dy}{dx} - y = x \] The integrating factor: \[ \mu(x) = e^{-\int 1 dx} = e^{-x} \] Multiplying the entire differential equation by \( e^{-x} \): \[ e^{-x}\frac{dy}{dx} - e^{-x} y = x e^{-x} \] Simplifies to: \[ \frac{d}{dx} (y e^{-x}) = x e^{-x} \] Integrate both sides: \[ y e^{-x} = \int x e^{-x} dx + C \] Using integration by parts for \( \int x e^{-x} dx \): - Let \( u = x \), \( dv = e^{-x} dx \) - Then \( du = dx \), \( v = -e^{-x} \) \[ \int x e^{-x} dx = -x e^{-x} + \int e^{-x} dx = -x e^{-x} - e^{-x} + C \] Thus, \[ y e^{-x} = -x e^{-x} - e^{-x} + C \] Multiply through by \( e^{x} \): \[ y = -x - 1 + C e^{x} \] Apply initial condition \( y(0) = 1 \): \[ 1 = -0 - 1 + C e^{0} \implies 1 = -1 + C \implies C = 2 \] **Final analytical solution:** \[ \boxed{ \boxed{ y(x) = -x - 1 + 2 e^{x} } } \] Calculate \( y(0.1) \): \[ y(0.1) = -0.1 - 1 + 2 e^{0.1} \] Calculate \( e^{0.1} \): \[ e^{0.1} \approx 1.10517 \] Therefore: \[ y(0.1) \approx -1.1 + 2 \times 1.10517 = -1.1 + 2.21034 = \boxed{1.11034} \] --- ## **2. Euler’s Method** ### **Initial data:** - \( x_0 = 0 \) - \( y_0 = 1 \) - \( h = 0.01 \) - Number of steps to reach \( x=0.1 \): \( n = \frac{0.1 - 0}{0.01} = 10 \) ### **Iteration formula:** \[ y_{n+1} = y_{n} + h \times f(x_n, y_n) \] where \( f(x, y) = x + y \). --- ### **Step-by-step calculations:** | Step (n) | \(x_n\) | \(y_n\) | \(f(x_n, y_n) = x_n + y_n\) | \(y_{n+1} = y_n + 0.01 \times f(x_n, y_n)\) | |---|---------|---------|------------------------------|------------------------------------------------| | 0 | 0.00 | 1.00000 | 0 + 1.00000 = 1.00000 | \(1.00000 + 0.01 \times 1.00000 = 1.01000\) | | 1 | 0.01 | 1.01000 | 0.01 + 1.01000 = 1.02000 | \(1.01000 + 0.01 \times 1.02000 = 1.02020\) | | 2 | 0.02 | 1.02020 | 0.02 + 1.02020 = 1.04020 | \(1.02020 + 0.01 \times 1.04020 = 1.03060\) | | 3 | 0.03 | 1.03060 | 0.03 + 1.03060 = 1.06060 | \(1.03060 + 0.01 \times 1.06060 = 1.04121\) | | 4 | 0.04 | 1.04121 | 0.04 + 1.04121 = 1.08121 | \(1.04121 + 0.01 \times 1.08121 = 1.05202\) | | 5 | 0.05 | 1.05202 | 0.05 + 1.05202 = 1.10202 | \(1.05202 + 0.01 \times 1.10202 = 1.06304\) | | 6 | 0.06 | 1.06304 | 0.06 + 1.06304 = 1.12304 | \(1.06304 + 0.01 \times 1.12304 = 1.07427\) | | 7 | 0.07 | 1.07427 | 0.07 + 1.07427 = 1.14427 | \(1.07427 + 0.01 \times 1.14427 = 1.08571\) | | 8 | 0.08 | 1.08571 | 0.08 + 1.08571 = 1.16571 | \(1.08571 + 0.01 \times 1.16571 = 1.09737\) | | 9 | 0.09 | 1.09737 | 0.09 + 1.09737 = 1.18737 | \(1.09737 + 0.01 \times 1.18737 = 1.10924\) | **Result at \( x=0.1 \):** \[ \boxed{ \textbf{Euler's method estimate:} \quad y(0.1) \approx \boxed{1.10924} } \] --- ## **3. Heun’s Method (Improved Euler)** ### **Iteration formula:** \[ y_{n+1} = y_n + \frac{h}{2} \left[ f(x_n, y_n) + f\left( x_{n+1}, y^*_n \right) \right] \] where the predictor: \[ y^*_n = y_n + h \times f(x_n, y_n) \] --- ### **Initial data:** Same as before: \( y_0=1 \), \( x_0=0 \), \( h=0.01 \). --- ### **Step-by-step calculations:** | Step (n) | \(x_n\) | \(y_n\) | \(f(x_n, y_n)\) | Predictor \( y^*_n \) | \(f(x_{n+1}, y^*_n)\) | \(y_{n+1}\) calculation | \(y_{n+1}\) value | |---|---------|---------|----------------|----------------------|------------------------|-------------------------|------------------| | 0 | 0.00 | 1.00000 | 1.00000 | \(1.00000 + 0.01 \times 1.00000 = 1.01000\) | \(0.01 + 1.01000 = 1.01100\) | \( y_1 = 1 + \frac{0.01}{2}(1 + 1.01100) \) | \( 1 + 0.005 \times 2.01100 = 1 + 0.010055 = 1.010055 \) | | 1 | 0.01 | 1.010055 | 0.01 + 1.010055 = 1.020055 | \(1.010055 + 0.01 \times 1.020055 = 1.020255\) | \(0.02 + 1.020255 = 1.030255\) | \( y_2 = 1.010055 + 0.005 \times (1.020055 + 1.030255) \) | \( 1.010055 + 0.005 \times 2.05031 = 1.010055 + 0.01025 = 1.020305 \) | | 2 | 0.02 | 1.020305 | 0.02 + 1.020305 = 1.040305 | \(1.020305 + 0.01 \times 1.040305 = 1.03061\) | \(0.03 + 1.03061 = 1.04061\) | \( y_3 = 1.020305 + 0.005 \times (1.040305 + 1.04061) \) | \( 1.020305 + 0.005 \times 2.080915 = 1.020305 + 0.0104 = 1.030705 \) | | 3 | 0.03 | 1.030705 | 0.03 + 1.030705 = 1.060705 | \(1.030705 + 0.01 \times 1.060705 = 1.04131\) | \(0.04 + 1.04131 = 1.08131\) | \( y_4 = 1.030705 + 0.005 \times (1.060705 + 1.08131) \) | \( 1.030705 + 0.005 \times 2.142015 = 1.030705 + 0.01071 = 1.041415 \) | | 4 | 0.04 | 1.041415 | 0.04 + 1.041415 = 1.081415 | \(1.041415 + 0.01 \times 1.081415 = 1.05223\) | \(0.05 + 1.05223 = 1.05223\) | \( y_5 = 1.041415 + 0.005 \times (1.081415 + 1.05223) \) | \( 1.041415 + 0.005 \times 2.13365 = 1.041415 + 0.01067 = 1.052085 \) | | 5 | 0.05 | 1.052085 | 0.05 + 1.052085 = 1.102085 | \(1.052085 + 0.01 \times 1.102085 = 1.06330\) | \(0.06 + 1.06330 = 1.06330\) | \( y_6 = 1.052085 + 0.005 \times (1.102085 + 1.06330) \) | \( 1.052085 + 0.005 \times 2.16539 = 1.052085 + 0.01083 = 1.062915 \) | | 6 | 0.06 | 1.062915 | 0.06 + 1.062915 = 1.122915 | \(1.062915 + 0.01 \times 1.122915 = 1.07352\) | \(0.07 + 1.07352 = 1.07352\) | \( y_7 = 1.062915 + 0.005 \times (1.122915 + 1.07352) \) | \( 1.062915 + 0.005 \times 2.19643 = 1.062915 + 0.01098 = 1.07389 \) | | 7 | 0.07 | 1.07389 | 0.07 + 1.07389 = 1.14389 | \(1.07389 + 0.01 \times 1.14389 = 1.08433\) | \(0.08 + 1.08433 = 1.08433\) | \( y_8 = 1.07389 + 0.005 \times (1.14389 + 1.08433) \) | \( 1.07389 + 0.005 \times 2.22822 = 1.07389 + 0.01114 = 1.08503 \) | | 8 | 0.08 | 1.08503 | 0.08 + 1.08503 = 1.16503 | \(1.08503 + 0.01 \times 1.16503 = 1.09718\) | \(0.09 + 1.09718 = 1.09718\) | \( y_9 = 1.08503 + 0.005 \times (1.16503 +

Certainly! Let's continue with the detailed calculations for Heun's method from where we left off. --- ### **Continuing Heun’s Method calculations:** **Step 8:** - \( x_8 = 0.08 \) - \( y_8 \approx 1.08503 \) Calculate \( f(x_8, y_8) \): \[ f(0.08, 1.08503) = 0.08 + 1.08503 = 1.16503 \] Predictor: \[ y^*_{8} = y_8 + h \times f(x_8, y_8) = 1.08503 + 0.01 \times 1.16503 = 1.08503 + 0.01165 = 1.09668 \] Next, evaluate \( f(x_9, y^*_{8}) \): \[ f(0.09, 1.09668) = 0.09 + 1.09668 = 1.18668 \] Update \( y_9 \): \[ y_9 = y_8 + \frac{h}{2} \left[ f(x_8, y_8) + f(x_9, y^*_{8}) \right] \] \[ y_9 = 1.08503 + 0.005 \times (1.16503 + 1.18668) = 1.08503 + 0.005 \times 2.35171 \] \[ y_9 = 1.08503 + 0.01176 = 1.09679 \] --- **Step 9:** - \( x_9 = 0.09 \) - \( y_9 \approx 1.09679 \) Calculate \( f(0.09, 1.09679) \): \[ f(0.09, 1.09679) = 0.09 + 1.09679 = 1.18679 \] Predictor: \[ y^*_{9} = y_9 + h \times f(0.09, y_9) = 1.09679 + 0.01 \times 1.18679 = 1.09679 + 0.01187 = 1.10866 \] Evaluate \( f(0.10, y^*_{9}) \): \[ f(0.10, 1.10866) = 0.10 + 1.10866 = 1.20866 \] Update \( y_{10} \): \[ y_{10} = y_9 + \frac{h}{2} \left[ f(0.09, y_9) + f(0.10, y^*_{9}) \right] \] \[ y_{10} = 1.09679 + 0.005 \times (1.18679 + 1.20866) = 1.09679 + 0.005 \times 2.39545 \] \[ y_{10} = 1.09679 + 0.01198 = \boxed{1.10877} \] --- ## **Final Results Summary:** | Method | Estimated \( y(0.1) \) | |-----------------|------------------------| | **Analytical** | **1.11034** | | **Euler** | **1.10924** | | **Heun** | **1.10877** | --- ## **Final Comments:** - The **analytical solution** gives \( y(0.1) \approx 1.11034 \). - The **Euler method** estimate is very close, slightly underestimating. - The **Heun’s method** offers an even more accurate estimate, very close to the analytical value. This demonstrates how Heun’s method improves upon Euler’s method by reducing the local truncation error, especially for small step sizes like \( h=0.01 \). --- **If you'd like further explanation or calculations for more steps, let me know!**

Certainly! Here's a summary of all the final answers for the problem: --- ### **1. Analytical Solution at \( x=0.1 \):** \[ y(0.1) = -0.1 - 1 + 2 e^{0.1} \approx -1.1 + 2 \times 1.10517 \approx 1.11034 \] --- ### **2. Euler’s Method estimate at \( x=0.1 \):** Using step size \( h=0.01 \): \[ \boxed{ \textbf{Euler’s estimate:} \quad y(0.1) \approx \mathbf{1.10924} } \] --- ### **3. Heun’s Method estimate at \( x=0.1 \):** Using step size \( h=0.01 \): \[ \boxed{ \textbf{Heun’s estimate:} \quad y(0.1) \approx \mathbf{1.10877} } \] --- ### **Final Comparison:** | Method | \( y(0.1) \) estimate | |-----------------|----------------------| | Analytical | **1.11034** | | Euler | **1.10924** | | Heun | **1.10877** | --- These are the detailed final answers for each method.

Certainly! Below is the complete, step-by-step detailed solution for the problem, including all parts of calculations written out explicitly without tables. --- ## **Given Data:** - Differential equation: \(\frac{dy}{dx} = x + y\) - Initial condition: \( y(0) = 1 \) - Step size: \( h = 0.01 \) - Target \( x \): \( 0.1 \) --- ## **Part 1: Analytical Solution** This is a linear first-order differential equation: \[ \frac{dy}{dx} - y = x \] **Step 1: Find the integrating factor (IF):** \[ \mu(x) = e^{-\int 1 dx} = e^{-x} \] **Step 2: Multiply the entire differential equation by \(\mu(x)\):** \[ e^{-x} \frac{dy}{dx} - e^{-x} y = x e^{-x} \] This simplifies to: \[ \frac{d}{dx} \left( y e^{-x} \right) = x e^{-x} \] **Step 3: Integrate both sides:** \[ y e^{-x} = \int x e^{-x} dx + C \] Apply integration by parts to evaluate \(\int x e^{-x} dx\): - Let \( u = x \), \( dv = e^{-x} dx \) - Then \( du = dx \), \( v = - e^{-x} \) \[ \int x e^{-x} dx = - x e^{-x} + \int e^{-x} dx = - x e^{-x} - e^{-x} + C \] **Step 4: Write the solution for \( y e^{-x} \):** \[ y e^{-x} = - x e^{-x} - e^{-x} + C \] Multiply through by \( e^{x} \): \[ y = - x - 1 + C e^{x} \] **Step 5: Find the constant \( C \) using initial condition \( y(0) = 1 \):** \[ 1 = - 0 - 1 + C e^{0} \implies 1 = -1 + C \implies C = 2 \] **Final analytical solution:** \[ \boxed{ y(x) = - x - 1 + 2 e^{x} } \] **Calculate \( y(0.1) \):** \[ y(0.1) = -0.1 - 1 + 2 e^{0.1} \] Since \( e^{0.1} \approx 1.10517 \): \[ y(0.1) \approx -1.1 + 2 \times 1.10517 = -1.1 + 2.21034 = \boxed{1.11034} \] --- ## **Part 2: Euler’s Method** We will approximate \( y(0.1) \) starting from \( y(0) = 1 \), with step size \( h=0.01 \). The formula is: \[ y_{n+1} = y_n + h \times f(x_n, y_n) \] where \( f(x, y) = x + y \). --- ### **Step-by-step calculation:** - **Step 0 (initial):** \[ x_0 = 0, \quad y_0 = 1 \] Calculate \( f(0, 1) \): \[ f(0, 1) = 0 + 1 = 1 \] Update \( y \): \[ y_1 = y_0 + 0.01 \times 1 = 1 + 0.01 = 1.01000 \] - **Step 1:** \[ x_1 = 0.01, \quad y_1 = 1.01000 \] Calculate \( f(0.01, 1.01000) \): \[ f(0.01, 1.01000) = 0.01 + 1.01000 = 1.02000 \] Update: \[ y_2 = y_1 + 0.01 \times 1.02000 = 1.01000 + 0.01020 = 1.02020 \] - **Step 2:** \[ x_2 = 0.02, \quad y_2 = 1.02020 \] Calculate \( f(0.02, 1.02020) \): \[ f(0.02, 1.02020) = 0.02 + 1.02020 = 1.04020 \] Update: \[ y_3 = 1.02020 + 0.01 \times 1.04020 = 1.02020 + 0.01040 = 1.03060 \] - **Step 3:** \[ x_3 = 0.03, \quad y_3 = 1.03060 \] Calculate \( f(0.03, 1.03060) \): \[ f(0.03, 1.03060) = 0.03 + 1.03060 = 1.06060 \] Update: \[ y_4 = 1.03060 + 0.01 \times 1.06060 = 1.03060 + 0.01061 = 1.04121 \] - **Step 4:** \[ x_4 = 0.04, \quad y_4 = 1.04121 \] Calculate \( f(0.04, 1.04121) \): \[ f(0.04, 1.04121) = 0.04 + 1.04121 = 1.08121 \] Update: \[ y_5 = 1.04121 + 0.01 \times 1.08121 = 1.04121 + 0.01081 = 1.05202 \] - **Step 5:** \[ x_5 = 0.05, \quad y_5=1.05202 \] Calculate \( f(0.05, 1.05202) \): \[ f(0.05, 1.05202) = 0.05 + 1.05202= 1.10202 \] Update: \[ y_6= 1.05202 + 0.01 \times 1.10202 = 1.05202 + 0.01102= 1.06304 \] - **Step 6:** \[ x_6= 0.06, \quad y_6= 1.06304 \] Calculate \( f(0.06, 1.06304) \): \[ f(0.06, 1.06304) = 0.06 + 1.06304= 1.12304 \] Update: \[ y_7= 1.06304 + 0.01 \times 1.12304= 1.06304 + 0.01123= 1.07427 \] - **Step 7:** \[ x_7= 0.07, \quad y_7= 1.07427 \] Calculate \( f(0.07, 1.07427) \): \[ f(0.07, 1.07427) = 0.07 + 1.07427= 1.14427 \] Update: \[ y_8= 1.07427 + 0.01 \times 1.14427= 1.07427 + 0.01144= 1.08571 \] - **Step 8:** \[ x_8= 0.08, \quad y_8= 1.08571 \] Calculate \( f(0.08, 1.08571) \): \[ f(0.08, 1.08571)= 0.08 + 1.08571= 1.16571 \] Update: \[ y_9= 1.08571 + 0.01 \times 1.16571= 1.08571 + 0.01166= 1.09737 \] - **Step 9 (final):** \[ x_9= 0.09, \quad y_9= 1.09737 \] Calculate \( f(0.09, 1.09737) \): \[ f(0.09, 1.09737)= 0.09 + 1.09737= 1.18737 \] Update: \[ y_{10} = y_9 + 0.01 \times 1.18737 = 1.09737 + 0.01187 = \boxed{1.10924} \] **Result from Euler's method at \( x=0.1 \):** \[ \boxed{ \text{Euler's estimate:} \quad y(0.1) \approx \mathbf{1.10924} } \] --- ## **Part 3: Heun’s Method (Improved Euler)** The formula: \[ y_{n+1} = y_n + \frac{h}{2} \left[ f(x_n, y_n) + f\left( x_{n+1}, y^*_n \right) \right] \] where \[ y^*_n = y_n + h \times f(x_n, y_n) \] --- ### **Step-by-step calculations:** - **Initial:** \[ x_0=0, \quad y_0=1 \] Calculate \( f(0,1) = 0 + 1= 1 \). Predictor: \[ y^*_0= y_0 + 0.01 \times 1= 1 + 0.01= 1.01000 \] Calculate \( f(0.01, y^*_0)= 0.01 + 1.01000= 1.02000 \). Update \( y_1 \): \[ y_1= y_0 + \frac{0.01}{2} (f(0,1)+f(0.01, y^*_0))= 1 + 0.005 (1 + 1.02000)= 1 + 0.005 \times 2.02000= 1 + 0.01010= 1.01010 \] - **Step 1:** \[ x_1=0.01, \quad y_1=1.01010 \] Calculate \( f(0.01, 1.01010)= 0.01 + 1.01010= 1.02010 \). Predict \( y^*_1 \): \[ y^*_1= y_1 + 0.01 \times 1.02010= 1.01010 + 0.01020= 1.02030 \] Calculate \( f(0.02, y^*_1)= 0.02 + 1.02030= 1.04030 \) Update \( y_2 \): \[ y_2= y_1 + \frac{0.01}{2} (1.02010 + 1.04030)= 1.01010 + 0.005 \times 2.06040= 1.01010 + 0.01030= 1.02040 \] - **Step 2:** \[ x_2=0.02, \quad y_2=1.02040 \] Calculate \( f(0.02, 1.02040)= 0.02 + 1.02040= 1.04040 \). Predict \( y^*_2 \): \[ y^*_2= 1.02040 + 0.01 \times 1.04040= 1.02040 + 0.01040= 1.03080 \] Calculate \( f(0.03, y^*_2)= 0.03 + 1.03080= 1.06080 \). Update \( y_3 \): \[ y_3= 1.02040 + 0.005 \times (1.04040 + 1.06080)= 1.02040 + 0.005 \times 2.10120= 1.02040 + 0.01051= 1.03091 \] - **Step 3:** \[ x_3=0.03, \quad y_3=1.03091 \] Calculate \( f(0.03, 1.03091)= 0.03 + 1.03091= 1.06091 \). Predict: \[ y^*_3= 1.03091 + 0.01 \times 1.06091= 1.03091 + 0.01061= 1.04152 \] Calculate: \[ f(0.04, y^*_3)= 0.04 + 1.04152= 1.08152 \] Update: \[ y_4= 1.03091 + 0.005 \times (1.06091 + 1.08152)= 1.03091 + 0.005 \times 2.14243= 1.03091 + 0.01071= 1.04162 \] - **Step 4:** \[ x

Certainly! Let's continue the detailed calculations for Heun’s method step-by-step from where we left off. --- ### **Step 4:** - Current \( x_4 = 0.04 \), current \( y_4 \approx 1.04162 \). Calculate \( f(0.04, y_4) \): \[ f(0.04, 1.04162) = 0.04 + 1.04162 = 1.08162 \] Predict \( y^*_4 \): \[ y^*_4 = y_4 + h \times f(0.04, y_4) = 1.04162 + 0.01 \times 1.08162 = 1.04162 + 0.01082 = 1.05244 \] Calculate \( f(0.05, y^*_4) \): \[ f(0.05, 1.05244) = 0.05 + 1.05244 = 1.10244 \] Update \( y_5 \): \[ y_5 = y_4 + \frac{h}{2} \left[ f(0.04, y_4) + f(0.05, y^*_4) \right] = 1.04162 + 0.005 \times (1.08162 + 1.10244) \] \[ = 1.04162 + 0.005 \times 2.18406 = 1.04162 + 0.01092 = 1.05254 \] --- ### **Step 5:** - \( x_5=0.05 \), \( y_5 \approx 1.05254 \). Calculate \( f(0.05, y_5) \): \[ f(0.05, 1.05254) = 0.05 + 1.05254 = 1.10254 \] Predict \( y^*_5 \): \[ y^*_5= 1.05254 + 0.01 \times 1.10254= 1.05254 + 0.01103= 1.06357 \] Calculate \( f(0.06, y^*_5) \): \[ f(0.06, 1.06357)= 0.06 + 1.06357= 1.12357 \] Update \( y_6 \): \[ y_6= 1.05254 + 0.005 \times (1.10254 + 1.12357)= 1.05254 + 0.005 \times 2.22611= 1.05254 + 0.01113= 1.06367 \] --- ### **Step 6:** - \( x_6=0.06 \), \( y_6 \approx 1.06367 \). Calculate \( f(0.06, y_6) \): \[ f(0.06, 1.06367)= 0.06 + 1.06367= 1.12367 \] Predict \( y^*_6 \): \[ y^*_6= 1.06367 + 0.01 \times 1.12367= 1.06367 + 0.01124= 1.07491 \] Calculate \( f(0.07, y^*_6) \): \[ f(0.07, 1.07491)= 0.07 + 1.07491= 1.14491 \] Update: \[ y_7= 1.06367 + 0.005 \times (1.12367 + 1.14491)= 1.06367 + 0.005 \times 2.26858= 1.06367 + 0.01134= 1.07501 \] --- ### **Step 7:** - \( x_7=0.07 \), \( y_7 \approx 1.07501 \). Calculate \( f(0.07, y_7) \): \[ f(0.07, 1.07501)= 0.07 + 1.07501= 1.14501 \] Predict \( y^*_7 \): \[ y^*_7= 1.07501 + 0.01 \times 1.14501= 1.07501 + 0.01145= 1.08646 \] Calculate \( f(0.08, y^*_7) \): \[ f(0.08, 1.08646)= 0.08 + 1.08646= 1.16646 \] Update: \[ y_8= 1.07501 + 0.005 \times (1.14501 + 1.16646)= 1.07501 + 0.005 \times 2.31147= 1.07501 + 0.01156= 1.08657 \] --- ### **Step 8:** - \( x_8=0.08 \), \( y_8 \approx 1.08657 \). Calculate \( f(0.08, y_8) \): \[ f(0.08, 1.08657)= 0.08 + 1.08657= 1.16657 \] Predict \( y^*_8 \): \[ y^*_8= 1.08657 + 0.01 \times 1.16657= 1.08657 + 0.01167= 1.09824 \] Calculate \( f(0.09, y^*_8) \): \[ f(0.09, 1.09824)= 0.09 + 1.09824= 1.18824 \] Update: \[ y_9= 1.08657 + 0.005 \times (1.16657 + 1.18824)= 1.08657 + 0.005 \times 2.35481= 1.08657 + 0.01177= 1.09834 \] --- ### **Step 9:** - \( x_9=0.09 \), \( y_9 \approx 1.09834 \). Calculate \( f(0.09, y_9) \): \[ f(0.09, 1.09834)= 0.09 + 1.09834= 1.18834 \] Predict \( y^*_9 \): \[ y^*_{9} = 1.09834 + 0.01 \times 1.18834= 1.09834 + 0.01188= 1.11022 \] Calculate \( f(0.10, y^*_{9}) \): \[ f(0.10, 1.11022)= 0.10 + 1.11022= 1.21022 \] Update \( y_{10} \): \[ y_{10} = 1.09834 + 0.005 \times (1.18834 + 1.21022) = 1.09834 + 0.005 \times 2.39856= 1.09834 + 0.01199= \boxed{1.11033} \] --- ## **Final Results:** - **Analytical solution at \( x=0.1 \):** \[ \boxed{ y(0.1) \approx 1.11034 } \] - **Euler’s method estimate:** \[ \boxed{ y(0.1) \approx 1.10924 } \] - **Heun’s method estimate:** \[ \boxed{ y(0.1) \approx 1.11033 } \] --- **Summary:** The analytical solution closely matches the Heun’s method estimate, with Euler’s method slightly underestimating. All values are very close due to the small step size.