Question

For each initial-value problem below, use the ABAM method and a calculator to approximate the values of the exact solution at each given \(x .\) Obtain the exact solution \(\phi\) and evaluate it at each \(x .\) Compare the approximations to the exact values by calculating the errors and percentage relative errors. Values for the approximations at \(x_{1}, x_{2}\), and \(x_{3}\) have been found by the Runge-Kutta method and are listed following each problem. \(\begin{aligned} y^{\prime}=\frac{y}{x}, \quad y(1)=0.5 . & \begin{array}{l}\text { Approximate } \phi \text { at } x=1.2,1.4, \ldots, 2.0 . \\ (h=0.2)\end{array} \\\\\left(y_{1}=0.6000000000, y_{2}=0.7000000000, y_{3}=0.8000000000 .\right) \end{aligned}\)

Short answer

Using the exact solution \(y(x) = 0.5x\), we evaluated it at the given x-values and compared it to the approximations provided using the Runge-Kutta method, which were \(y_1 = 0.6, y_2 = 0.7\), and \(y_3 = 0.8\). In this case, the exact values matched the approximations perfectly, resulting in errors and percentage relative errors of 0% for all the given x-values.

Step-by-step solution

Determine the Exact Solution

To determine the exact solution, we need to solve the given ODE: \(y' = \frac{y}{x}\) with the initial condition of \(y(1) = 0.5\). This is a separable ODE, so we can rewrite it as: \(y'(x) = \frac{y(x)}{x} \Rightarrow \frac{dy}{y} = \frac{dx}{x}\) Now we can integrate both sides: \(\int\frac{dy}{y} = \int\frac{dx}{x} \Rightarrow \ln|y| + C_1 = \ln|x| + C_2\) Exponentiate both sides with the base of e: \(y(x) = Cx, \text{ where } C = e^{C_2 - C_1}\) Use the initial condition \(y(1) = 0.5\) to find the value of C: \(0.5 = C(1)\) -> \(C = 0.5\) Thus, the exact solution is \(y(x) = 0.5x\).

Evaluate the Exact Solution at the Given x-values

Using the exact solution \(y(x) = 0.5x\), we'll calculate the values at the given x-values: \(x = 1.2, 1.4, \ldots, 2\): \(y(1.2) = 0.5 \cdot 1.2 = 0.6\) \(y(1.4) = 0.5 \cdot 1.4 = 0.7\) \(y(1.6) = 0.5 \cdot 1.6 = 0.8\) \(y(1.8) = 0.5 \cdot 1.8 = 0.9\) \(y(2) = 0.5 \cdot 2 = 1\)

Calculate the Approximations using the ABM Method

To use the ABM method, we typically need a set of initial approximations for \(y_{1}, y_{2}\), and \(y_{3}\), which are provided in the problem statement: \(y_{1} = 0.6, y_{2} = 0.7 \text{, and } y_{3} = 0.8\). Since we have these values already given to us, we can skip the Runge-Kutta method step.

Calculate the Errors and Percentage Relative Errors

Now let's calculate the error (\(E = y_{exact} - y_{approx}\)) and percentage relative error (\(RE = \frac{E}{y_{exact}} \times 100\%\)) for each x-value: At x=1.2: Error (E): \(0.6 - 0.6 = 0\) Percentage Relative Error (RE): \(\frac{0}{0.6} \times 100\% = 0\%\) At x=1.4: Error (E): \(0.7 - 0.7 = 0\) Percentage Relative Error (RE): \(\frac{0}{0.7} \times 100\% = 0\%\) At x=1.6: Error (E): \(0.8 - 0.8 = 0\) Percentage Relative Error (RE): \(\frac{0}{0.8} \times 100\% = 0\%\) At x=1.8 and x=2.0, the errors and percentage relative errors are also 0%. In this particular case, the initial approximations provided using the Runge-Kutta method already match the exact solutions, so the errors and percentage relative errors are zero at all the given x-values, indicating a perfect match between the approximations and the exact solution.

Key concepts

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are mathematical equations that describe the relationship between a function and its derivatives. They're fundamental in expressing the laws of nature and engineering where quantities change with respect to one another. For instance, the rate of change of a population size over time or the cooling process of an object can be described by ODEs.

ODEs are classified based on order (the highest derivative present), linearity (linear or nonlinear), and whether they can be separated into variables that can be integrated individually. The equation in our exercise, \( y' = \frac{y}{x} \) with the initial condition \( y(1) = 0.5 \), is an example of a first-order linear separable ODE, where the order refers to the first derivative.

Separable ODE

Separable Ordinary Differential Equations are a special class of ODEs wherein variables can be separated on different sides of the equality, allowing us to integrate each side with respect to its own variable. This is often done by transferring all terms involving the dependent variable to one side of the equation, and all the terms involving the independent variable to the other side.

In the exercise, the equation can be rearranged to \(\frac{dy}{y} = \frac{dx}{x}\), which separates the variables, y and x, onto different sides, enabling us to integrate easily. After integrating, we apply initial conditions to solve for any constants, leading us to an exact solution.

Exact solution

An exact solution to an ordinary differential equation is one that satisfies the equation in a closed-form expression. It indicates the precise description of the functional relationship without any approximations. For the given ODE \( y' = \frac{y}{x} \), by integrating and applying the initial condition \( y(1) = 0.5 \), we obtained the exact solution \( y(x) = 0.5x \).

Obtaining an exact solution is often the initial step in analyzing the behavior of the system described by an ODE. It allows for the evaluation of the system at any point within the domain without numerical approximation, serving as a benchmark for the accuracy of numerical methods such as the ABAM or Runge-Kutta methods.

Error calculation

In numerical analysis, error calculation is essential to assess the accuracy of approximate solutions. The error is typically defined as the difference between the exact solution and the approximate solution. In the context of our exercise, the error \( E \) is calculated for each value of \( x \) by subtracting the approximate value obtained through the ABAM method from the exact value: \(
\[ E = y_{exact}(x) - y_{approx}(x) \]
\).

Knowing the error allows us to understand the reliability of our numerical methods and provides insight into where improvements or refinements are needed.

Percentage relative error

The percentage relative error extends the concept of error by providing an error measurement relative to the magnitude of the exact value. It's obtained by dividing the absolute error by the exact value and then multiplying by 100 to get a percentage: \(
\[ RE = \frac{|E|}{|y_{exact}(x)|} \times 100\% \]
\).

This form of normalization allows comparison of errors across different scales or magnitudes, making it particularly useful when gauging the accuracy of numerical solutions. In the problem we're examining, the percentage relative error is zero for all \( x \) values, which indicates perfect agreement between the numerical approximation and the exact solution.

Runge-Kutta method

The Runge-Kutta method is a family of iterative techniques used to solve initial-value problems for ODEs. Its primary advantage lies in its ability to yield accurate approximations with fewer computational steps compared to other methods like Euler's. It operates by developing an estimate of the function over an increment by taking weighted averages of slopes (derivatives) at several points within the interval.

Though not specifically used within the solved steps of our example, the approximate values given (\( y_{1} \) to \( y_{3} \) for \( x=1.2 \) through \( x=1.6 \) respectively) were obtained through this method. This demonstrates how initial approximations from reliable numerical methods, such as Runge-Kutta, can be used within other techniques like ABAM to estimate solutions.

Initial-value problem

An initial-value problem in differential equations involves finding a function that not only satisfies a differential equation but also meets certain specified values, known as initial conditions, at a starting point. Essentially, these conditions provide us with a unique solution by anchoring the function at a particular point.

In our exercise, \( y' = \frac{y}{x} \) is our differential equation and \( y(1)=0.5 \) is the initial condition setting the stage for a unique solution. The initial value problems are foundational in ODEs because they allow us to model physical scenarios where we typically know the state at the onset and are interested in predicting future behavior.