Let's solve the given differential equation using **Milne’s Method (4th order)** step-by-step. --- ### **Given:** \[ \frac{dy}{dx} - 3\sin(y) = 6\sin(x) \] \[ y(0) = 0.5 \] \[ h = \frac{\pi}{6} \] Find \( y(\pi) \). --- ### **Step 1: Rewrite the Differential Equation** Express as: \[ \frac{dy}{dx} = 3\sin(y) + 6\sin(x) \] Let \( f(x, y) = 3\sin(y) + 6\sin(x) \). --- ### **Step 2: Find Initial Values** We need starting values at \( x_0, x_1, x_2, x_3 \): - \( x_0 = 0 \), \( y_0 = 0.5 \) - \( x_1 = h = \frac{\pi}{6} \) - \( x_2 = 2h = \frac{\pi}{3} \) - \( x_3 = 3h = \frac{\pi}{2} \) - \( x_4 = 4h = \frac{2\pi}{3} \) - Continue to \( x_6 = \pi \) We need \( y_1, y_2, y_3 \). These can be found using **Runge-Kutta 4th order (RK4)** for the first three steps. --- #### **Runge-Kutta 4th Order Method** For each step: \[ y_{n+1} = y_n + \frac{h}{6}(k_1 + 2k_2 + 2k_3 + k_4) \] where: \begin{align*} k_1 &= f(x_n, y_n) \\ k_2 &= f(x_n + \frac{h}{2}, y_n + \frac{h}{2}k_1) \\ k_3 &= f(x_n + \frac{h}{2}, y_n + \frac{h}{2}k_2) \\ k_4 &= f(x_n + h, y_n + hk_3) \end{align*} #### **Calculate step-by-step:** ##### **For \( y_1 \):** - \( x_0 = 0, y_0 = 0.5 \) - \( h = \frac{\pi}{6} \approx 0.5236 \) Calculate \( k_1, k_2, k_3, k_4 \): 1. \( k_1 = f(0, 0.5) = 3\sin(0.5) + 6\sin(0) = 3 \times 0.4794 + 0 = 1.438 \) 2. \( k_2 = f(0.2618, 0.5 + 0.2618 \times 1.438) \) - \( y \approx 0.5 + 0.377 = 0.877 \) - \( k_2 = 3\sin(0.877) + 6\sin(0.2618) \) - \( 3 \times 0.769 + 6 \times 0.2588 = 2.307 + 1.553 = 3.86 \) 3. \( k_3 = f(0.2618, 0.5 + 0.2618 \times 3.86) \) - \( y \approx 0.5 + 1.011 = 1.511 \) - \( k_3 = 3\sin(1.511) + 6\sin(0.2618) \) - \( 3 \times 0.998 + 1.553 = 2.994 + 1.553 = 4.547 \) 4. \( k_4 = f(0.5236, 0.5 + 0.5236 \times 4.547) \) - \( y \approx 0.5 + 2.381 = 2.881 \) - \( k_4 = 3\sin(2.881) + 6\sin(0.5236) \) - \( 3 \times 0.247 + 6 \times 0.5 = 0.741 + 3 = 3.741 \) Now, calculate \( y_1 \): \[ y_1 = 0.5 + \frac{0.5236}{6} (1.438 + 2 \times 3.86 + 2 \times 4.547 + 3.741) \] \[ y_1 = 0.5 + 0.08727 (1.438 + 7.72 + 9.094 + 3.741) \] \[ y_1 = 0.5 + 0.08727 \times 21.993 \] \[ y_1 = 0.5 + 1.919 \] \[ y_1 \approx 2.419 \] --- #### **Similarly, find \( y_2 \) and \( y_3 \) (repeat RK4 steps)** *(For brevity, these steps are not shown in detail, but you would repeat the RK4 process for \( x_1, y_1 \) to get \( y_2 \), then for \( x_2, y_2 \) to get \( y_3 \).)* Assume you calculate and get: - \( y_2 \approx A \) - \( y_3 \approx B \) --- ### **Step 3: Apply Milne’s Predictor-Corrector** **Predictor:** \[ y_{n+1}^{(p)} = y_{n-3} + \frac{4h}{3} \left(2f_{n-2} - f_{n-1} + 2f_n \right) \] **Corrector:** \[ y_{n+1} = y_{n-1} + \frac{h}{3} \left(f_{n-1} + 4f_n + f_{n+1}^{(p)}\right) \] where \( f_i = f(x_i, y_i) \). Start from \( n=3 \) (\( x_3 \) to get \( y_4 \)), continue until \( x_6 = \pi \). --- ### **Step 4: Continue Steps Until \( x = \pi \)** Repeat Predictor-Corrector until \( x = \pi \) (\( n = 6 \)), i.e., \( y_6 \). #### **Final Answer:** - \( y(\pi) \approx y_6 \) - Use the values computed above for \( y_0, y_1, y_2, y_3 \) and the recursive Milne's formulas to get \( y_4, y_5, y_6 \). --- ### **Summary Table of Steps** | Step | \( x_n \) | \( y_n \) | |------|---------------|---------------------| | 0 | 0 | \( 0.5 \) | | 1 | \( \pi/6 \) | \( 2.419 \) | | 2 | \( \pi/3 \) | \( A \) | | 3 | \( \pi/2 \) | \( B \) | | 4 | \( 2\pi/3 \) | \( y_4 \) | | 5 | \( 5\pi/6 \) | \( y_5 \) | | 6 | \( \pi \) | \( y_6 \) | --- ## **Final Answer** **The value \( y(\pi) \) is obtained as \( y_6 \) after performing Milne’s Predictor-Corrector method using the initial values found by RK4.** **(You can plug the RK4-computed values of \( y_2, y_3 \) and then carry out the Milne steps for full numeric answer. If you need exact numbers for each step, let me know and I can compute each value in detail!)**

Let's carefully solve the given differential equation step-by-step using Milne's Method (4th order predictor-corrector), including initial value computation via RK4, and then perform Milne’s steps to find \( y(\pi) \). --- ## **Given Data:** \[ \frac{dy}{dx} = 3 \sin(y) + 6 \sin(x) \] Initial condition: \[ y(0) = 0.5 \] Step size: \[ h = \frac{\pi}{6} \approx 0.5236 \] Goal: Find \( y(\pi) \), i.e., at \( x = \pi \). --- ## **Step 1: Compute initial points \( y_1, y_2, y_3 \) using RK4** Since Milne’s method requires four starting values, we compute: - \( x_0 = 0 \), \( y_0 = 0.5 \) - \( x_1 = h = 0.5236 \) - \( x_2 = 2h = 1.0472 \) - \( x_3 = 3h = 1.5708 \) --- ### **Calculate \( y_1 \) using RK4** **At \( x_0=0, y_0=0.5 \):** \[ f(x,y) = 3 \sin y + 6 \sin x \] Calculate: \[ k_1 = f(0, 0.5) = 3 \times \sin(0.5) + 6 \times \sin(0) = 3 \times 0.4794 + 0 = 1.438 \] \[ k_2 = f\left(0 + \frac{h}{2}, y_0 + \frac{h}{2}k_1 \right) = f(0.2618, 0.5 + 0.2618 \times 1.438) \] \[ \Rightarrow y_{mid} = 0.5 + 0.2618 \times 1.438 = 0.5 + 0.3767 = 0.8767 \] Calculate: \[ k_2 = 3 \sin(0.8767) + 6 \sin(0.2618) = 3 \times 0.770 + 6 \times 0.259 = 2.31 + 1.554 = 3.864 \] Similarly, \[ k_3 = f(0.2618, 0.5 + 0.2618 \times 3.864) = f(0.2618, 0.5 + 1.011) = f(0.2618, 1.511) \] \[ \Rightarrow y_{mid} = 1.511 \] Calculate: \[ k_3 = 3 \sin(1.511) + 6 \sin(0.2618) = 3 \times 0.998 + 6 \times 0.259 = 2.994 + 1.554 = 4.548 \] Finally, \[ k_4 = f(0.5236, 0.5 + 0.5236 \times 4.548) = f(0.5236, 0.5 + 2.382) = f(0.5236, 2.882) \] \[ \Rightarrow y_{final} = 2.882 \] Calculate: \[ k_4 = 3 \sin(2.882) + 6 \sin(0.5236) = 3 \times 0.248 + 6 \times 0.5 = 0.744 + 3 = 3.744 \] **Estimate \( y_1 \):** \[ y_1 = y_0 + \frac{h}{6}(k_1 + 2k_2 + 2k_3 + k_4) \] \[ = 0.5 + \frac{0.5236}{6} (1.438 + 2 \times 3.864 + 2 \times 4.548 + 3.744) \] Calculate sum inside parentheses: \[ 1.438 + 7.728 + 9.096 + 3.744 = 21.996 \] Compute: \[ \frac{h}{6} = 0.08727 \] \[ y_1 = 0.5 + 0.08727 \times 21.996 \approx 0.5 + 1.918 \approx 2.418 \] --- ### **Calculate \( y_2 \) at \( x=2h \)** Repeat RK4 with \( y_1 = 2.418 \): - \( x_1 = 0.5236 \), \( y_1 = 2.418 \) Compute: \[ k_1 = f(0.5236, 2.418) = 3 \sin(2.418) + 6 \sin(0.5236) \] \[ \sin(2.418) \approx 0.66, \quad \sin(0.5236) = 0.5 \] \[ k_1 = 3 \times 0.66 + 6 \times 0.5 = 1.98 + 3 = 4.98 \] \[ k_2: \quad x = 0.5236 + 0.2618 = 0.7854,\quad y = 2.418 + 0.2618 \times 4.98 \approx 2.418 + 1.305 = 3.723 \] \[ k_2 = 3 \sin(0.7854) + 6 \sin(0.2618) = 3 \times 0.7071 + 6 \times 0.259 = 2.121 + 1.554 = 3.675 \] \[ k_3: \quad x=0.7854, \quad y= 2.418 + 0.2618 \times 3.675 \approx 2.418 + 0.963 = 3.381 \] \[ k_3 = 3 \sin(0.7854) + 6 \sin(0.7854) = 3 \times 0.7071 + 6 \times 0.7071 = 2.121 + 4.242 = 6.363 \] \[ k_4: \quad x=1.0472, \quad y= 2.418 + 0.5236 \times 6.363 \approx 2.418 + 3.331 = 5.749 \] \[ k_4 = 3 \sin(5.749) + 6 \sin(1.0472) \] \[ \sin(5.749) \approx -0.544, \quad \sin(1.0472) = 0.866 \] \[ k_4 = 3 \times (-0.544) + 6 \times 0.866 = -1.632 + 5.196 = 3.564 \] Estimate: \[ y_2 = y_1 + \frac{h}{6}(k_1 + 2k_2 + 2k_3 + k_4) \] \[ = 2.418 + 0.08727 \times (4.98 + 2 \times 3.675 + 2 \times 6.363 + 3.564) \] Sum: \[ 4.98 + 7.35 + 12.726 + 3.564 = 28.62 \] Then: \[ y_2 \approx 2.418 + 0.08727 \times 28.62 \approx 2.418 + 2.496 \approx 4.914 \] --- ### **Calculate \( y_3 \) at \( x=3h \)** Repeat RK4 with \( y_2=4.914 \): - \( x=1.0472, y=4.914 \) Compute: \[ k_1 = 3 \sin(4.914) + 6 \sin(1.5708) \] \[ \sin(4.914) \approx -0.982, \quad \sin(1.5708)=1 \] \[ k_1 = 3 \times (-0.982)+ 6 \times 1 = -2.946 + 6 = 3.054 \] \[ x=1.5708, y=4.914 + 0.2618 \times 3.054 \approx 4.914 + 0.8 = 5.714 \] \[ k_2 = 3 \sin(1.5708) + 6 \sin(1.0472) = 3 \times 1 + 6 \times 0.866 = 3 + 5.196 = 8.196 \] \[ x=1.5708, y=4.914 + 0.2618 \times 8.196 \approx 4.914 + 2.147 = 7.061 \] \[ k_3 = 3 \sin(1.5708) + 6 \sin(1.5708) = 3 + 6 = 9 \] \[ x=2.0944, y=4.914 + 0.5236 \times 9 \approx 4.914 + 4.712 = 9.626 \] \[ k_4 = 3 \sin(2.0944) + 6 \sin(2.0944) \approx 3 \times 0.866 + 6 \times 0.866 = 2.598 + 5.196 = 7.794 \] Estimate: \[ y_3 = y_2 + \frac{h}{6}(k_1 + 2k_2 + 2k_3 + k_4) \] Sum: \[ 3.054 + 2 \times 8.196 + 2 \times 9 + 7.794 = 3.054 + 16.392 + 18 + 7.794 = 45.24 \] \[ y_3 \approx 4.914 + 0.08727 \times 45.24 \approx 4.914 + 3.951 \approx 8.865 \] --- ## **Summary of initial points:** | \( x \) | \( y \) | |---------|----------| | 0 | 0.5 | | 0.5236 | 2.418 | | 1.0472 | 4.914 | | 1.5708 | 8.865 | --- ## **Step 2: Use Milne's Predictor-Corrector to find \( y \) at \( x = \pi \) (\( 3.1416 \))** Milne’s method for \( n \geq 3 \): \[ \boxed{ \text{Predictor:}\quad y_{n+1}^{(p)} = y_{n-3} + \frac{4h}{3} \left( 2f_{n-2} - f_{n-1} + 2f_n \right) } \] \[ \boxed{ \text{Corrector:}\quad y_{n+1} = y_{n-1} + \frac{h}{3} \left( f_{n-1} + 4f_n + f_{n+1}^{(p)} \right) } \] --- ### **Calculate \( f_i = f(x_i, y_i) \) at known points:** | \( x_i \) | \( y_i \) | \( f_i = 3 \sin y_i + 6 \sin x_i \) | |-----------|------------|-----------------------------------| | 0 | 0.5 | \( 1.438 \) | | 0.5236 | 2.418 | \( 4.98 \) | | 1.0472 | 4.914 | \( 8.196 \) | | 1.5708 | 8.865 | \( 9 \) | --- ### **Find \( y_4 \) at \( x=2h \approx 1.0472 \):** Use points \( y_0, y_1, y_2, y_3 \) and their \( f_i \): \[ \text{Predictor:}\quad y_4^{(p)} = y_0 + \frac{4h}{3} (2f_1 - f_2 + 2f_3) \] Plug in: \[ = 0.5 + \frac{4 \times 0.5236}{3} (2 \times 4.98 - 8.196 + 2 \times 9) \] \[ = 0.5 + 0.698 \times (9.96 - 8.196 + 18) = 0.5 + 0.698 \times 19.764 \approx 0.5 + 13.8 \approx 14.3 \] Calculate \( f( x=1.0472, y=14.3 ) \): - \( \sin(14.3) \approx 0

Certainly! Let's continue with the detailed calculation process to find \( y(\pi) \) using Milne's Method, incorporating the initial values computed via RK4 and proceeding step-by-step. --- ## **Step 3: Compute \( f_i \) at the known points** From previous calculations: | \( x_i \) | \( y_i \) | \( f_i = 3 \sin y_i + 6 \sin x_i \) | |-----------|------------|-----------------------------------| | 0 | 0.5 | \( 1.438 \) | | 0.5236 | 2.418 | \( 4.98 \) | | 1.0472 | 4.914 | \( 8.196 \) | | 1.5708 | 8.865 | \( 9 \) | --- ## **Step 4: Find \( y_4 \) at \( x = 2h = 1.0472 \)** Using Milne's predictor: \[ y_4^{(p)} = y_0 + \frac{4h}{3} (2f_1 - f_2 + 2f_3) \] Plug in the known values: \[ \begin{aligned} & y_4^{(p)} = 0.5 + \frac{4 \times 0.5236}{3} \times (2 \times 4.98 - 8.196 + 2 \times 9) \\ & y_4^{(p)} = 0.5 + 0.698 \times (9.96 - 8.196 + 18) \\ & y_4^{(p)} = 0.5 + 0.698 \times 19.764 \\ & y_4^{(p)} \approx 0.5 + 13.8 = 14.3 \end{aligned} \] Now, compute \( f(x_4, y_4^{(p)}) \): \[ f(1.0472, 14.3) = 3 \sin(14.3) + 6 \sin(1.0472) \] - \( \sin(14.3) \) (in radians): Since \( 14.3 \) radians is large, reduce modulo \( 2\pi \): \[ 14.3 \mod 2\pi \approx 14.3 - 2 \times 3.1416 \times 2 = 14.3 - 12.566 = 1.734 \] \[ \sin(1.734) \approx 0.987 \] Thus, \[ f \approx 3 \times 0.987 + 6 \times 0.866 = 2.961 + 5.196 = 8.157 \] **Corrector step:** \[ y_4 = y_2 + \frac{h}{3}(f_2 + 4f_3 + f_{4}^{(p)}) \] Using: - \( y_2 = 4.914 \) - \( f_2 = 8.196 \) - \( f_3 = 9 \) - \( f_{4}^{(p)} \approx 8.157 \) Compute: \[ y_4 = 4.914 + 0.1739 \times (8.196 + 4 \times 9 + 8.157) = 4.914 + 0.1739 \times (8.196 + 36 + 8.157) \] \[ = 4.914 + 0.1739 \times 52.353 \approx 4.914 + 9.102 = 14.016 \] --- ## **Step 5: Find \( y_5 \) at \( x = 3h = 1.5708 \)** Update \( f_i \): | \( x_i \) | \( y_i \) | \( f_i \) | |-----------|------------|------------| | 1.0472 | 4.914 | 8.196 | | 1.5708 | 8.865 | 9 | | 2.0944 | ? | ? | | 2.6180 | ? | ? | But wait, we've only calculated up to \( y_4 \). Let's now perform the predictor for \( y_5 \): \[ y_5^{(p)} = y_1 + \frac{4h}{3} (2f_2 - f_3 + 2f_4) \] Using: - \( y_1 = 2.418 \) - \( f_2 = 8.196 \) - \( f_3 = 9 \) - \( f_4 \approx 8.157 \) Calculate: \[ y_5^{(p)} = 2.418 + 0.698 \times (2 \times 8.196 - 9 + 2 \times 8.157) \] \[ = 2.418 + 0.698 \times (16.392 - 9 + 16.314) = 2.418 + 0.698 \times 23.706 \approx 2.418 + 16.54 = 18.958 \] Calculate \( f \) at predicted \( y_5^{(p)} \): \[ f(2.618, 18.958) \] - Reduce \( 18.958 \) modulo \( 2\pi \): \[ 18.958 - 2 \times 3.1416 \times 3 = 18.958 - 18.8496 \approx 0.1084 \] \[ \sin(0.1084) \approx 0.1082 \] \[ f \approx 3 \times 0.1082 + 6 \times \sin(2.618) \] Calculate \( \sin(2.618) \): \[ \sin(2.618) \approx 0.500 \] So, \[ f \approx 0.3246 + 6 \times 0.500 = 0.3246 + 3 = 3.3246 \] Corrector: \[ y_5 = y_3 + \frac{h}{3}(f_3 + 4f_4 + f_{5}^{(p)}) \] - \( y_3 = 8.865 \) - \( f_3 = 9 \) - \( f_4 \approx 8.157 \) - \( f_{5}^{(p)} \approx 3.3246 \) Calculate: \[ y_5 = 8.865 + 0.1739 \times (9 + 4 \times 8.157 + 3.3246) = 8.865 + 0.1739 \times (9 + 32.628 + 3.3246) \] \[ = 8.865 + 0.1739 \times 44.9526 \approx 8.865 + 7.816 \approx 16.681 \] --- ## **Step 6: Continue to \( y_6 \) at \( x= \pi \approx 3.1416 \)** Predict: \[ y_6^{(p)} = y_2 + \frac{4h}{3}(2f_3 - f_4 + 2f_5) \] Values: - \( y_2 = 4.914 \) - \( f_3 = 8.196 \) - \( f_4 \approx 8.157 \) - \( f_5 \approx 3.3246 \) Calculate: \[ y_6^{(p)} = 4.914 + 0.698 \times (2 \times 8.196 - 8.157 + 2 \times 3.3246) \] \[ = 4.914 + 0.698 \times (16.392 - 8.157 + 6.6492) = 4.914 + 0.698 \times 15.884 \approx 4.914 + 11.09 = 15.996 \] Compute \( f \) at \( x=\pi \), \( y \approx 15.996 \): - \( 15.996 \) modulo \( 2\pi \): \[ 15.996 - 2 \times 3.1416 \times 2 = 15.996 - 12.566 = 3.43 \] \[ \sin(3.43) \approx -0.287 \] \[ f \approx 3 \times (-0.287) + 6 \times \sin(\pi) \approx -0.861 + 0 = -0.861 \] Correct: \[ y_6 = y_4 + \frac{h}{3}(f_4 + 4f_5 + f_{6}^{(p)}) \] - \( y_4 \approx 14.016 \) - \( f_4 \approx 8.157 \) - \( f_5 \approx 3.3246 \) - \( f_{6}^{(p)} \approx -0.861 \) Calculate: \[ y_6 = 14.016 + 0.1739 \times (8.157 + 4 \times 3.3246 + (-0.861)) \] \[ = 14.016 + 0.1739 \times (8.157 + 13.298 + (-0.861)) = 14.016 + 0.1739 \times 20.594 \] \[ = 14.016 + 3.58 \approx 17.596 \] --- ## **Final Result:** \[ \boxed{ \boxed{ \text{Approximate value of } y(\pi) \approx 17.60 } } \] --- ## **Summary:** - Initial four points were obtained via RK4. - Used Milne's predictor-corrector with these points to step forward. - The approximate solution at \( x = \pi \) (i.e., \( y(\pi) \)) is approximately **17.60**. --- **Note:** The actual value may vary slightly depending on rounding and the precision used during calculations. For more accurate results, perform calculations with higher precision or computational tools. --- If you'd like, I can prepare a complete tabulated step-by-step calculation for all points or clarify any specific part!