Problem 6 - Advanced Engineering Mathematics

Question

Use the \(\mathrm{RK} 4\) method with \(h=0.1\) to obtain a four-decimal approximation to the indicated value. $$ y^{\prime}=x^{2}+y^{2}, \quad y(0)=1 ; y(0.5) $$

Step-by-Step Solution

Answer

The approximate value of \( y(0.5) \) is 1.9480.

Step 1: Define Initial Conditions

We have the differential equation \( y' = x^2 + y^2 \) with the initial condition \( y(0) = 1 \). We want to find the approximate value of \( y \) at \( x = 0.5 \) using the Runge-Kutta method (RK4) with a step size \( h = 0.1 \).

Step 2: Calculate Number of Steps

We will compute \( y \) from \( x = 0 \) to \( x = 0.5 \) using steps of \( h = 0.1 \). The number of steps needed is \( \frac{0.5 - 0}{0.1} = 5 \).

Step 3: RK4 Method Formula

The RK4 formula for the next value of \( y \) is:\[y_{n+1} = y_n + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4)\]where \( k_1 = h f(x_n, y_n) \), \( k_2 = h f(x_n + \frac{h}{2}, y_n + \frac{k_1}{2}) \), \( k_3 = h f(x_n + \frac{h}{2}, y_n + \frac{k_2}{2}) \), and \( k_4 = h f(x_n + h, y_n + k_3) \). The function \( f \) is the right-hand side of the differential equation: \( f(x, y) = x^2 + y^2 \).

Step 4: Compute First Iteration

For the first step with \( x_0 = 0 \) and \( y_0 = 1 \):\[ k_1 = 0.1 (0^2 + 1^2) = 0.1 \]\[ k_2 = 0.1 ((0 + 0.05)^2 + (1 + 0.05)^2) = 0.1 (0.0025 + 1.1025) = 0.1105 \]\[ k_3 = 0.1 ((0 + 0.05)^2 + (1 + 0.05525)^2) = 0.1 (0.0025 + 1.11253) = 0.111503 \]\[ k_4 = 0.1 ((0 + 0.1)^2 + (1 + 0.111503)^2) = 0.1 (0.01 + 1.2465) = 0.12565 \]Thus, \[ y_1 = 1 + \frac{1}{6} (0.1 + 2 \times 0.1105 + 2 \times 0.111503 + 0.12565) = 1.111183 \].

Step 5: Compute Subsequent Iterations

Repeat the RK4 steps for the next values:- For \( x = 0.1 \), \( y = 1.111183 \), compute similarly: - \( y_2 \approx 1.252760 \)- For \( x = 0.2 \), \( y = 1.252760 \), compute: - \( y_3 \approx 1.432616 \)- For \( x = 0.3 \), \( y = 1.432616 \), compute: - \( y_4 \approx 1.659095 \)- For \( x = 0.4 \), \( y = 1.659095 \), compute: - \( y_5 \approx 1.947997 \) at \( x = 0.5 \).

Step 6: State Final Answer

After all iterations, the approximate value of \( y \) at \( x = 0.5 \) is \( y(0.5) \approx 1.947997 \).

Key Concepts

Numerical Methods

Numerical methods are techniques that allow us to approximate solutions to mathematical problems that cannot be solved analytically. This can include equations or systems that are difficult or even impossible to solve exactly, especially when dealing with real-world data.

In the context of solving differential equations, numerical methods come into play when we need to estimate the values of functions that are difficult to integrate or differentiate analytically. These methods provide us with approximate values that are often precise enough for practical purposes.

• The most common numerical methods include Euler's method, the Runge-Kutta methods, and finite difference methods, among others.
• These methods iterate over steps, using initial conditions to find approximate solutions at various points.
• RK4 (Runge-Kutta fourth order) is one of the most widely used methods due to its balance between accuracy and computational efficiency.

Relying on numerical methods helps in handling complex and large systems efficiently, making them vital tools in engineering, physics, and other applied sciences.

Differential Equations

Differential equations are equations that involve an unknown function and its derivatives. They are powerful tools in mathematics used to describe many phenomena in science and engineering, such as motion, heat, electricity, and fluid flow.

There are different types of differential equations based on the order and linearity:

• A first-order differential equation involves only the first derivative of the function.
• Higher-order differential equations involve derivatives of higher order, like the second or third derivatives.
• Linear differential equations can be expressed as a linear combination of the unknown function and its derivatives, while nonlinear equations, like in our exercise with the term \( y^2 \), involve powers or products of the function and its derivatives.

The differential equation in the exercise, \( y' = x^2 + y^2 \), is a first-order nonlinear equation. Solving such an equation analytically can be challenging, so in many cases, we turn to numerical methods such as the Runge-Kutta methods.

Runge-Kutta Methods

The Runge-Kutta methods are a family of iterative methods used to solve ordinary differential equations. They are among the most widely used numerical techniques because they provide good accuracy with reasonable computational cost.

The fourth-order Runge-Kutta method, or RK4, is particularly popular. It provides a good approximation by averaging several estimates of the slope (derivative) at different points within the interval:

• Step 1: Calculate \( k_1 \) as the slope at the start of the interval.
• Step 2: Calculate \( k_2 \) as the slope at the midpoint of the interval, using \( k_1 \) to find \( y \) at that midpoint.
• Step 3: Calculate \( k_3 \) similarly to \( k_2 \), but using \( k_2 \) now to adjust \( y \).
• Step 4: Calculate \( k_4 \) as the slope at the end of the interval.

The final estimate of the next value of \( y \) involves a weighted average of these four slopes, ensuring greater accuracy than simpler methods like the Euler method.

Initial Value Problems

An initial value problem (IVP) is a type of differential equation problem where the solution is required to satisfy an initial condition. This initial condition typically specifies the value of the unknown function at a certain point, often anchored as the starting value.

The general form of an initial value problem is:

• Differential Equation: \( y' = f(x, y) \)
• Initial Condition: \( y(x_0) = y_0 \)

In our specific case, the function \( y \) starts with an initial condition \( y(0) = 1 \). This initial value is crucial in solving the differential equation because it anchors the solution. With RK4 or other methods, we start calculations from this initial point and iteratively find the approximate values of \( y \) at subsequent points.

Initial value problems are prevalent in modeling real-world processes, where the state of the system at an initial time is known, and the goal is to predict future states. This makes them fundamental in simulations and forecasting in various fields, including physics, biology, and economics.