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 $$ y^{\prime}=\frac{y^{2}+x y-x^{2}}{x^{2}}, \quad y(2)=-1 $$ at \(x=1.1,1.2,1.3, \ldots 2.0 .\) Compare these approximate values with the values of the exact solution $$ y=\frac{x\left(4-3 x^{2}\right)}{4+3 x^{2}} $$ which can be obtained by referring to Example \(2.4 .3 .\)

Short answer

In this exercise, we applied the Runge-Kutta method of order 4 (RK4) to find approximate values of the given differential equation for three different step sizes (\(h=0.1, h=0.05,\) and \(h=0.025\)). After obtaining the approximate solutions, we calculated the exact solution and compared the error between the exact and approximate values. In general, the Runge-Kutta method is very accurate for solving differential equations, and as the step size decreases, the accuracy improves. By comparing the errors, we can observe that smaller step sizes result in a more accurate approximation, but also require more computation.

Step-by-step solution

Define the Given Function

Let's define the given function denoted by f(x, y): $$ f(x, y) = \frac{y^{2}+x y-x^{2}}{x^{2}} $$ Now, the differential equation given is: $$ y^{\prime}(x) = f(x, y(x)) $$ with the initial condition \(y(2)=-1\).

Implement the Runge-Kutta Method

Now we will implement the Runge-Kutta method of order 4 (RK4) to find the approximate values of the solution. RK4 method formula is given by: $$ \begin{aligned} &k_1 = h*f(x_n, y_n) \\ &k_2 = h*f(x_n + \frac{h}{2}, y_n + \frac{k_1}{2}) \\ &k_3 = h*f(x_n + \frac{h}{2}, y_n + \frac{k_2}{2}) \\ &k_4 = h*f(x_n + h, y_n + k_3) \\ &y_{n+1} = y_n + \frac{k_1 + 2*k_2 + 2*k_3 + k_4}{6} \end{aligned} $$ Here, \(h\) is the step size, and \({(x_n, y_n)}\) represents the n-th point in the numerical solution. For each step size, we will iterate these steps and find the approximate values of \(y\) at the given \(x\) values.

Apply RK4 for Each Step Size

Using the RK4 method for each step size \(h=0.1\), \(h=0.05\), and \(h=0.025\), we will find the solution values for \(x = 1.1, 1.2, 1.3, \ldots, 2.0\). Write down the solutions for each case.

Compute the Exact Solution

Calculate the values for the exact solution given by the formula: $$ y=\frac{x\left(4-3 x^{2}\right)}{4+3 x^{2}} $$ for \(x = 1.1, 1.2, 1.3, \ldots, 2.0\).

Compare Approximate Values with the Exact Solution

Now, compare the Runge-Kutta solution values obtained for each step size with the exact solution values. You can compute the error by finding the absolute difference between the approximate values and the exact solution. This will give you an idea of how accurate the Runge-Kutta method is for each step size.

Key concepts

Numerical Methods for Differential Equations

When solving differential equations, analytical solutions are often difficult or impossible to find. This is where numerical methods come into play. These methods provide a way to estimate solutions of differential equations, allowing us to solve problems that are otherwise too complex.
One popular numerical method that is often used is the Runge-Kutta method. It's effective in estimating the solutions to differential equations with high accuracy. Instead of striving for exact answers, Runge-Kutta offers approximations that become more precise with smaller step sizes.
Numerical methods are crucial for engineers, scientists, and mathematicians because they can solve complex systems where analytical solutions don't exist.

• Provide approximate solutions where exact solutions aren't possible
• Used for modeling real-world phenomena
• Offer different levels of complexity and accuracy

Initial Value Problems

Initial value problems (IVPs) are a specific type of differential equation problem where the solution must satisfy a given condition at the start, known as the initial condition. An example of an initial condition is specifying the value of the function at a particular point, such as given by the condition in the exercise, \(y(2) = -1\).
This type of problem is critical in many scientific applications where the future state depends on the current or initial state. These problems are typically solved using numerical methods like the Runge-Kutta method.
The initial condition ensures that the solution to the differential equation is unique. Without it, there could be multiple solutions that satisfy the equation.

• Defines the starting point of the solution
• Ensures a unique solution
• Common in simulations and scientific computations

Exact vs Approximate Solutions

Differential equations often require solutions that are either exact or approximate. An exact solution is a precise answer that solves the differential equation fully, as given by the specific formula \(y=\frac{x\left(4-3x^{2}\right)}{4+3x^{2}}\).
However, not every differential equation can be solved exactly, which is why approximate solutions are critical. Using methods like Runge-Kutta, we can calculate approximate solutions that are close to the exact solution but within an acceptable range of error.
The accuracy of these approximations depends on factors such as the numerical method used and the step size chosen, which in turn can be assessed by comparing approximate values to the exact ones.

• Exact solutions solve the equation precisely
• Approximate solutions are estimates, often close to exact
• Comparison between the two helps assess accuracy

Step Size and Accuracy

Step size is a vital parameter in numerical methods like the Runge-Kutta method. It signifies how much we advance the independent variable each step when calculating the approximate solution. Smaller step sizes usually lead to more accurate results because they reduce the approximation error.
However, smaller steps also mean more calculations, which can be time-consuming. In the exercise, different step sizes \(h=0.1\), \(h=0.05\), and \(h=0.025\) were used to illustrate how they affect accuracy. As the step size decreases, the approximations become closer to the exact solution.
Choosing an appropriate step size involves balancing accuracy and computational effort.

• Smaller step sizes often yield more accurate results
• Require more computational resources
• A balance between step size and computation is essential