Problem 4 - Advanced Engineering Mathematics

Question

Use the RK4 method with \(h=0.1\) to obtain a four-decimal approximation to the indicated value. $$ y^{\prime}=4 x-2 y, \quad y(0)=2 ; y(0.5) $$

Step-by-Step Solution

Answer

The value of \( y(0.5) \) is approximately 1.1022.

Step 1: Initialize Parameters

The task is to approximate the value of \( y(0.5) \) using the RK4 method with a step size \( h=0.1 \). The initial condition is \( y(0) = 2 \).

Step 2: Define the Function

The differential equation is given as \( y' = 4x - 2y \), so the function for the RK4 method is \( f(x, y) = 4x - 2y \).

Step 3: RK4 Formula Overview

The RK4 method computes the next value \( y_{i+1} \) using:\[y_{i+1} = y_i + \frac{h}{6} (k_1 + 2k_2 + 2k_3 + k_4)\]where:\[\begin{align*}k_1 & = f(x_i, y_i), \k_2 & = f(x_i + \frac{h}{2}, y_i + \frac{h}{2} k_1), \k_3 & = f(x_i + \frac{h}{2}, y_i + \frac{h}{2} k_2), \k_4 & = f(x_i + h, y_i + hk_3).\end{align*}\]

Step 4: First Iteration: Calculate at \( x = 0 \)

With \( x_0 = 0 \), \( y_0 = 2 \): \[ \begin{align*}k_1 & = f(0, 2) = 4(0) - 2(2) = -4, \k_2 & = f(0.05, 2 - 0.05(4)) = 4(0.05) - 2(1.8) = -3.1, \k_3 & = f(0.05, 2 - 0.05(3.1)) = 4(0.05) - 2(1.845) = -3.14, \k_4 & = f(0.1, 2 + 0.1(-3.14)) = 4(0.1) - 2(1.686) = -2.972.\end{align*}\]Then:\[ y_1 = 2 + \frac{0.1}{6}(-4 + 2(-3.1) + 2(-3.14) - 2.972) = 1.6863.\]

Step 5: Second Iteration: Calculate at \( x = 0.1 \)

With \( x_1 = 0.1 \), \( y_1 = 1.6863 \): \[ \begin{align*}k_1 & = f(0.1, 1.6863) = 4(0.1) - 2(1.6863) = -2.9726, \k_2 & = f(0.15, 1.6863 + 0.05(-2.9726)) = 4(0.15) - 2(1.5377) = -2.6255, \k_3 & = f(0.15, 1.6863 + 0.05(-2.6255)) = 4(0.15) - 2(1.5550) = -2.645, \k_4 & = f(0.2, 1.6863 + 0.1(-2.645)) = 4(0.2) - 2(1.4218) = -2.2436.\end{align*}\]Then:\[ y_2 = 1.6863 + \frac{0.1}{6}(-2.9726 + 2(-2.6255) + 2(-2.645) -2.2436) = 1.4218.\]

Step 6: Third Iteration: Calculate at \( x = 0.2 \)

With \( x_2 = 0.2 \), \( y_2 = 1.4218 \): \[ \begin{align*}k_1 & = f(0.2, 1.4218) = 4(0.2) - 2(1.4218) = -2.2436, \k_2 & = f(0.25, 1.4218 + 0.05(-2.2436)) = 4(0.25) - 2(1.30964) = -1.86928, \k_3 & = f(0.25, 1.4218 + 0.05(-1.86928)) = 4(0.25) - 2(1.328336) = -1.906672, \k_4 & = f(0.3, 1.4218 + 0.1(-1.906672)) = 4(0.3) - 2(1.2311328) = -1.4622656.\end{align*}\]Then:\[ y_3 = 1.4218 + \frac{0.1}{6}(-2.2436 + 2(-1.86928) + 2(-1.906672) -1.4622656) = 1.2311.\]

Step 7: Fourth Iteration: Calculate at \( x = 0.3 \)

With \( x_3 = 0.3 \), \( y_3 = 1.2311 \): \[ \begin{align*}k_1 & = f(0.3, 1.2311) = 4(0.3) - 2(1.2311) = -1.4622, \k_2 & = f(0.35, 1.2311 + 0.05(-1.4622)) = 4(0.35) - 2(1.15899) = -1.01798, \k_3 & = f(0.35, 1.2311 + 0.05(-1.01798)) = 4(0.35) - 2(1.180201) = -1.060402, \k_4 & = f(0.4, 1.2311 + 0.1(-1.060402)) = 4(0.4) - 2(1.1250598) = -0.65012.\end{align*}\]Then:\[ y_4 = 1.2311 + \frac{0.1}{6}(-1.4622 + 2(-1.01798) + 2(-1.060402) -0.65012) = 1.1251.\]

Step 8: Fifth Iteration: Calculate at \( x = 0.4 \)

With \( x_4 = 0.4 \), \( y_4 = 1.1251 \): \[ \begin{align*}k_1 & = f(0.4, 1.1251) = 4(0.4) - 2(1.1251) = -0.6502, \k_2 & = f(0.45, 1.1251 + 0.05(-0.6502)) = 4(0.45) - 2(1.09259) = -0.18518, \k_3 & = f(0.45, 1.1251 + 0.05(-0.18518)) = 4(0.45) - 2(1.115841) = -0.231682, \k_4 & = f(0.5, 1.1251 + 0.1(-0.231682)) = 4(0.5) - 2(1.1019318) = 0.1961364.\end{align*}\]Then:\[ y_5 = 1.1251 + \frac{0.1}{6}(-0.6502 + 2(-0.18518) + 2(-0.231682) + 0.1961364) = 1.1022.\]

Step 9: Obtain Approximation for \( y(0.5) \)

After five iterations using the RK4 method, we approximate \( y(0.5) \) as \( 1.1022 \).

Key Concepts

Numerical Methods

Numerical methods are techniques used to find approximate solutions to complex mathematical problems. They play a critical role in solving equations that cannot be tackled analytically. Numerical methods help us approximate values like integrals, roots of equations, or solutions to differential equations. In the given exercise, the Runge-Kutta method, specifically the fourth-order version (RK4), is a type of numerical method. It is used for solving ordinary differential equations (ODEs).

The RK4 method is particularly popular because:

• It provides a good balance between accuracy and computational complexity.
• It is more precise compared to simpler methods like Euler's method.
• It is widely employed in engineering and physics due to its robustness.

By using a step-by-step process, the RK4 method estimates values at successive points and refines its accuracy with each iteration.

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. These equations model a wide array of phenomena in nature, such as heat transfer, population growth, and motion. In the exercise, the differential equation is expressed as \( y' = 4x - 2y \). Often, these equations describe how a quantity changes over time or space.

Key points to understand about differential equations:

• They provide relationships between functions and their rates of change.
• Solving them gives us the nature of the variation of physical quantities.
• Some differential equations, due to their complexity, require numerical solutions rather than analytical approaches.

In this solution, the Runge-Kutta method iteratively approximates the function \( y \) over small steps, providing a practical approach to solving the differential equation numerically.

Initial Value Problems

Initial value problems (IVPs) are a type of differential equation where the solution is sought from an initial starting point. An IVP consists of a differential equation along with a specified value at a particular point, called the initial condition. In our exercise, the initial value is \( y(0) = 2 \).

Understanding IVPs involves:

• Recognizing that the initial condition provides a starting context for solving the differential equation.
• Using the initial value to uniquely determine the solution curve from numerous possibilities.
• Ensuring proper numerical methods are utilized to track the solution as it evolves.

The role of the RK4 method here is pivotal, as it uses the initial condition to kickstart the iterative process, ensuring that subsequent calculations yield an accurate approximation of the solution at desired points. This accurate progression from the initial point is why numerical methods are crucial to solving IVPs.