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1Assignment Problems
srsohatmspmomeone YF 2Y Ex Eh. P00) Sd onan y vi second ores Rige.
Kutta with h=0.2 2nd y(0)=1
b. Using the previous results use Adam-Bashforthipredictor) two step explicit method in conjunction with Adam-
Moulton two step implicit method (corrector) to approximate y (0.04) with h=0.01
. Use MAT

VIPSolutions · Accepted Answer

Let's solve each part step by step, with explanations at each step:

---

### **a. Solve the ODE using 2nd order Runge-Kutta with $ h = 0.2 $, $ y(0) = 1 $, and approximate $ y(0.4) $**

#### **Step 1: Write the ODE in standard form**

Given:
$$
y' + 2y = x + 4, \quad y(0) = 1
$$
Rewrite:
$$
y' = x + 4 - 2y
$$
Let $ f(x, y) = x + 4 - 2y $.

---

#### **Step 2: Apply the 2nd order Runge-Kutta (Heun's Method)**

RK2 (Heun's):
$$
k_1 = f(x_n, y_n)
$$
$$
k_2 = f(x_n + h, y_n + h k_1)
$$
$$
y_{n+1} = y_n + \frac{h}{2}(k_1 + k_2)
$$

**At $ x_0 = 0, y_0 = 1, h = 0.2 $:**

**First step ($ x_0 = 0 \rightarrow x_1 = 0.2 $):**
- $ k_1 = f(0, 1) = 0 + 4 - 2(1) = 2 $
- $ k_2 = f(0.2, 1 + 0.2 \times 2) = f(0.2, 1.4) = 0.2 + 4 - 2(1.4) = 4.2 - 2.8 = 1.4 $
- $ y_1 = 1 + \frac{0.2}{2}(2 + 1.4) = 1 + 0.1 \times 3.4 = 1 + 0.34 = 1.34 $

**Second step ($ x_1 = 0.2 \rightarrow x_2 = 0.4 $):**
- $ k_1 = f(0.2, 1.34) = 0.2 + 4 - 2(1.34) = 4.2 - 2.68 = 1.52 $
- $ k_2 = f(0.4, 1.34 + 0.2 \times 1.52) = f(0.4, 1.34 + 0.304) = f(0.4, 1.644) = 0.4 + 4 - 2(1.644) = 4.4 - 3.288 = 1.112 $
- $ y_2 = 1.34 + \frac{0.2}{2}(1.52 + 1.112) = 1.34 + 0.1 \times 2.632 = 1.34 + 0.2632 = 1.6032 $

**Explanation:**  
We calculated the increments using the RK2 method for each step and updated the $ y $ value accordingly.

---

**Final Answer for (a):**  
$$
\boxed{y(0.4) \approx 1.6032}
$$

---

### **b. Use Adams-Bashforth (predictor) and Adams-Moulton (corrector) to approximate $ y(0.04) $ with $ h = 0.01 $**
*For this, let's use the notation $ y_n, y_{n-1} $ at $ x_n, x_{n-1} $. We need two initial values, so use RK2 for the first two steps.*

#### **Step 1: Get $ y_0 $ and $ y_1 $ using RK2 with $ h = 0.01 $**
- $ y_0 = 1 $
- $ k_1 = f(0, 1) = 2 $
- $ k_2 = f(0.01, 1 + 0.01 \times 2) = f(0.01, 1.02) = 0.01 + 4 - 2(1.02) = 4.01 - 2.04 = 1.97 $
- $ y_1 = 1 + \frac{0.01}{2}(2 + 1.97) = 1 + 0.005 \times 3.97 = 1 + 0.01985 = 1.01985 $

**Explanation:**  
We use RK2 for the first step to get $ y_1 $ at $ x = 0.01 $.

---

#### **Step 2: Apply Adam-Bashforth (predictor) and Adam-Moulton (corrector) for $ y_2 $ at $ x_2 = 0.02 $, $ y_3 $ at $ x_3 = 0.03 $, $ y_4 $ at $ x_4 = 0.04 $**

**Adams-Bashforth 2-step (predictor):**
$$
y_{n+1}^{(p)} = y_n + \frac{h}{2}[3f(x_n, y_n) - f(x_{n-1}, y_{n-1})]
$$
**Adams-Moulton 2-step (corrector):**
$$
y_{n+1} = y_n + \frac{h}{2}[f(x_{n+1}, y_{n+1}^{(p)}) + f(x_n, y_n)]
$$

Let's do calculations for $ y_2 $ ($ x_2 = 0.02 $):
- $ f(x_0, y_0) = f(0, 1) = 2 $
- $ f(x_1, y_1) = f(0.01, 1.01985) = 0.01 + 4 - 2(1.01985) = 4.01 - 2.0397 = 1.9703 $

**Predictor:**
$$
y_2^{(p)} = y_1 + \frac{0.01}{2} [3 \times 1.9703 - 2] = 1.01985 + 0.005 \times (5.9109 - 2) = 1.01985 + 0.005 \times 3.9109 = 1.01985 + 0.0195545 = 1.0394045
$$

**Corrector:**
$ f(x_2, y_2^{(p)}) = f(0.02, 1.0394045) = 0.02 + 4 - 2 \times 1.0394045 = 4.02 - 2.078809 = 1.941191 $
$$
y_2 = y_1 + \frac{0.01}{2} [1.9703 + 1.941191] = 1.01985 + 0.005 \times 3.911491 = 1.01985 + 0.0195575 = 1.0394075
$$
Repeat similarly for $ y_3 $ and $ y_4 $:

**For $ y_3 $:**
- $ f(x_2, y_2) = f(0.02, 1.0394075) = 1.941185 $
- Predictor:
  $$
  y_3^{(p)} = y_2 + 0.005 [3 \times 1.941185 - 1.9703] = 1.0394075 + 0.005 \times (5.823555 - 1.9703) = 1.0394075 + 0.005 \times 3.853255 = 1.0394075 + 0.0192663 = 1.0586738
  $$
- Corrector:
  $ f(x_3, y_3^{(p)}) = f(0.03, 1.0586738) = 0.03 + 4 - 2 \times 1.0586738 = 4.03 - 2.1173476 = 1.912652 $
  $$
  y_3 = 1.0394075 + 0.005 \times (1.941185 + 1.912652) = 1.0394075 + 0.005 \times 3.853837 = 1.0394075 + 0.019269 = 1.0586765
  $$

**For $ y_4 $:**
- $ f(x_3, y_3) = f(0.03, 1.0586765) = 1.912647 $
- Predictor:
  $$
  y_4^{(p)} = 1.0586765 + 0.005 [3 \times 1.912647 - 1.941185] = 1.0586765 + 0.005 \times (5.737941 - 1.941185) = 1.0586765 + 0.005 \times 3.796756 = 1.0586765 + 0.0189838 = 1.0776603
  $$
- Corrector:
  $ f(x_4, y_4^{(p)}) = f(0.04, 1.0776603) = 0.04 + 4 - 2 \times 1.0776603 = 4.04 - 2.1553206 = 1.884679 $
  $$
  y_4 = 1.0586765 + 0.005 \times (1.912647 + 1.884679) = 1.0586765 + 0.005 \times 3.797326 = 1.0586765 + 0.0189866 = 1.0776631
  $$

**Explanation:**  
We used RK2 for the first step, and then applied the Adams-Bashforth predictor and Adams-Moulton corrector for successive steps.

---

**Final Answer for (b):**
$$
\boxed{y(0.04) \approx 1.0777}
$$

---

### **c. MATLAB Code for RK2 from $ x = 0 $ to $ x = 1 $, $ h = 0.2 $, $ y(0) = 1 $:**

```matlab
h = 0.2;
x = 0:h:1;
y = zeros(size(x));
y(1) = 1; % initial condition

f = @(x, y) x + 4 - 2*y;

for n = 1:length(x)-1
    k1 = f(x(n), y(n));
    k2 = f(x(n) + h, y(n) + h*k1);
    y(n+1) = y(n) + h/2*(k1 + k2);
end

plot(x, y, '-o')
xlabel('x')
ylabel('y')
title('RK2 Approximation of y'' + 2y = x + 4, y(0)=1')
grid on
```

---

**Summary of Answers:**

**a.** $ y(0.4) \approx 1.6032 $ (RK2, $ h = 0.2 $)  
**b.** $ y(0.04) \approx 1.0777 $ (Adams-Bashforth/Moulton, \( h = 0

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