Question

Use the Runge-Kutta method with step sizes \(h=0.1, h=0.05,\) and \(h=0.025\) to find approximate values of the solution of $$ y^{\prime}+\frac{1}{x} y=\frac{7}{x^{2}}+3, \quad y(1)=\frac{3}{2} $$ at \(x=0.5,0.6, \ldots, 1.5\). Compare these approximate values with the values of the exact solution $$ y=\frac{7 \ln x}{x}+\frac{3 x}{2} $$ which can be obtained by the method discussed in Section 2.1 .

Short answer

#tag_title#Short Answer: The differential equation is given as \(y'(x) = -\frac{1}{x}y(x) + \frac{7}{x^2} + 3\) with the exact solution formula \(y(x) = \frac{7 \ln x}{x}+\frac{3x}{2}\). To find the approximate solution using the Runge-Kutta method, we will iterate using different step sizes (h = 0.1, 0.05, 0.025) for x values from 0.5 to 1.5. By comparing the approximate values obtained from the Runge-Kutta method to the exact solution values, we can observe the difference in accuracy between each step size.

Step-by-step solution

Define the given differential equation

The given differential equation can be written as $$ y'(x) = -\frac{1}{x}y(x) + \frac{7}{x^2} + 3 $$ with the initial condition \(y(1) = \frac{3}{2}\).

Implement the Runge-Kutta method using a step size h

To implement the Runge-Kutta method, we shall use the following formulas: k1 = f(x_n, y_n) k2 = f(x_n + h/2, y_n + h*k1/2) k3 = f(x_n + h/2, y_n + h*k2/2) k4 = f(x_n + h, y_n + h*k3) y_(n+1) = y_n + h*(k1 + 2*k2 + 2*k3 + k4)/6 x_(n+1) = x_n + h Here, f(x, y) is the function defined in the previous step, h is the step size, and x_n, y_n are the current x and y values.

Calculate approximate values for each given x using h = 0.1, 0.05, 0.025

Iteratively apply the Runge-Kutta method three times for each step size, until you reach a x value of 1.5. For \(h = 0.1\): Start at x = 1 and y = 3/2. Continue the iteration from step 2 until you reach x = 1.5. For \(h = 0.05\): Start at x = 1 and y = 3/2. Continue the iteration from step 2 until you reach x = 1.5. For \(h = 0.025\): Start at x = 1 and y = 3/2. Continue the iteration from step 2 until you reach x = 1.5.

Calculate the exact values using the given formula

For each x value (from 0.5 to 1.5 with increments of 0.1), use the exact solution formula to calculate y value: Exact solution: \(y = \frac{7 \ln x}{x}+\frac{3x}{2}\)

Compare approximate Runge-Kutta values with exact values

For each x value, compare the approximate Runge-Kutta values (calculated in step 3) for all step sizes (h = 0.1, 0.05, 0.025) with the exact values (calculated in step 4) and observe the differences. The comparison should be in this format: - x value - Approximate values using Runge-Kutta method with step sizes h=0.1, h=0.05, h=0.025 - Exact values using given solution formula - Difference between approximate and exact values for each step size (h=0.1, h=0.05, h=0.025) This will allow you to see how the accuracy of the Runge-Kutta method changes with the step size and how close the approximate values are to the exact values for each x value.

Key concepts

Differential Equations

Differential equations are a cornerstone in understanding phenomena that involve rates of change in various fields such as physics, engineering, and economics. At their core, differential equations are mathematical equations that relate some function with its derivatives. In real-world applications, these functions often represent physical quantities like velocity, concentration, or temperature, and their derivatives can be understood as the rates of change of these quantities.

Differential equations come in many forms, with varying complexity, from simple separable equations to more challenging nonlinear systems. The equation given in the exercise, \(y' + \frac{1}{x} y = \frac{7}{x^{2}} + 3\), is a first-order linear ordinary differential equation (ODE). The 'first-order' indicates it involves only the first derivative of the function \(y(x)\), and 'linear' signifies that the equation is a linear combination of the function and its derivatives. Solving these equations analytically can sometimes provide an 'exact' solution, a closed-form expression that holds at every point within the domain of the problem.

Numerical Methods

When an exact solution to a differential equation is not readily available or is too complex to determine, numerical methods come to the rescue. These methods convert the problem of solving differential equations into a form that can be readily computed using algorithms and computers. Runge-Kutta methods, including the one used in the exercise, are among the most popular numerical techniques for solving initial value problems for ODEs.

The Runge-Kutta methods approximate the solution by employing a series of 'predictor-corrector' steps that systematically reduce error. Specifically, these methods calculate the next value of the dependent variable (in our case, \(y\)) based on the current value and the rate of change (given by the differential equation). The 'fourth-order' Runge-Kutta method, which the exercise implements, strikes a balance between accuracy and computational intensity. It involves calculating four intermediate slopes (\(k1, k2, k3, k4\)) to estimate the new point. Larger step sizes generally lead to less accuracy, prompting practitioners to use smaller steps for more accurate results, as seen in the step size variation \(h = 0.1, 0.05, 0.025\) in the example.

Exact Solution Comparison

Comparing numerical approximations to the exact solution is crucial in assessing the performance and reliability of a numerical method like Runge-Kutta. When an exact analytical solution is available, as it is with the given function \(y = \frac{7 \ln x}{x} + \frac{3 x}{2}\), it becomes the standard against which we measure the approximations.

In the exercise, the comparison process involves calculating the approximation error at various points, highlighting how numerical solutions can deviate from the exact values. The error is the difference between the approximated value and the exact solution, and depending on the context, several metrics can be used to quantify this error. By comparing values obtained with different step sizes, one can observe that a smaller step size typically yields an approximation closer to the exact value, indicating a more accurate method. However, it is also important to recognize that smaller step sizes mean more computational work. Therefore, striking the right balance between accuracy and computational efficiency is key to solving real-world problems using numerical methods.