Problem 7 - Advanced Engineering Mathematics

Question

Use the Runge-Kutta method to approximate \(x(0.2)\) and \(y(0.2) .\) First use \(h=0.2\) and then use \(h=0.1\). Use a numerical solver and \(h=0.1\) to graph the solution in a neighborhood of \(t=0\). $$ \begin{aligned} &x^{\prime}=2 x-y \\ &y^{\prime}=x \\ &x(0)=6, y(0)=2 \end{aligned} $$

Step-by-Step Solution

Answer

For \(h=0.2\), \(x(0.2) \approx 8.372\), \(y(0.2) \approx 3.314\); for \(h=0.1\), \(x(0.2) \approx 8.222\), \(y(0.2) \approx 3.296\).

Step 1: Preparing Runge-Kutta Formulas

The Runge-Kutta method involves the calculation of intermediate values in order to compute the next point. For two equations and derivatives \( x'(t) = f(x, y) = 2x - y \) and \( y'(t) = g(x, y) = x \), we calculate using the following formulas: \[ k_1 = h f(x_n, y_n), \; l_1 = h g(x_n, y_n) \] \[ k_2 = h f(x_n + \frac{1}{2}k_1, y_n + \frac{1}{2}l_1), \; l_2 = h g(x_n + \frac{1}{2}k_1, y_n + \frac{1}{2}l_1) \] \[ k_3 = h f(x_n + \frac{1}{2}k_2, y_n + \frac{1}{2}l_2), \; l_3 = h g(x_n + \frac{1}{2}k_2, y_n + \frac{1}{2}l_2) \] \[ k_4 = h f(x_n + k_3, y_n + l_3), \; l_4 = h g(x_n + k_3, y_n + l_3) \] Then, update to \( x_{n+1} \) and \( y_{n+1} \) using: \[ x_{n+1} = x_n + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4) \] \[ y_{n+1} = y_n + \frac{1}{6}(l_1 + 2l_2 + 2l_3 + l_4) \]

Step 2: Applying Runge-Kutta with h=0.2

Start with initial conditions \( x_0 = 6 \), \( y_0 = 2 \), and step size \( h = 0.2 \). Calculate the intermediate values \( k_1, l_1, k_2, l_2, k_3, l_3, k_4, l_4 \) for the first step \((t = 0) \rightarrow (t = 0.2)\):\[ k_1 = 0.2 \times (2 \times 6 - 2) = 2 \]\[ l_1 = 0.2 \times 6 = 1.2 \]\[ k_2 = 0.2 \times (2 \times (6 + 1) - (2 + 0.6)) = 2.24 \]\[ l_2 = 0.2 \times (6 + 0.6) = 1.32 \]\[ k_3 = 0.2 \times (2 \times (6 + 1.12) - (2 + 0.66)) = 2.312 \]\[ l_3 = 0.2 \times (6 + 0.66) = 1.332 \] \[ k_4 = 0.2 \times (2 \times (6 + 1.332) - (2 + 1.332)) = 2.60 \]\[ l_4 = 0.2 \times (6 + 1.332) = 1.666 \] Update the values: \[ x(0.2) = 6 + \frac{1}{6}(2 + 2 \times 2.24 + 2 \times 2.312 + 2.60) \approx 8.372 \] \[ y(0.2) = 2 + \frac{1}{6}(1.2 + 2 \times 1.32 + 2 \times 1.332 + 1.666) \approx 3.314 \]

Step 3: Applying Runge-Kutta with h=0.1

We repeat the process using \( h = 0.1 \). Start with \( x_0 = 6 \), \( y_0 = 2 \) and calculate \( k_1, l_1, k_2, l_2, k_3, l_3, k_4, l_4 \) for two steps from \( t = 0 \rightarrow t = 0.1 \rightarrow t = 0.2 \). First step for \( t = 0 \rightarrow t = 0.1 \):\[ k_1 = 0.1 \times (2 \times 6 - 2) = 1 \]\[ l_1 = 0.1 \times 6 = 0.6 \]\[ k_2 = 0.1 \times (2 \times (6 + 0.5) - (2 + 0.3)) = 1.12 \]\[ l_2 = 0.1 \times (6 + 0.3) = 0.63 \]\[ k_3 = 0.1 \times (2 \times (6 + 0.56) - (2 + 0.315)) = 1.141 \]\[ l_3 = 0.1 \times (6 + 0.315) = 0.6315 \] \[ k_4 = 0.1 \times (2 \times (6 + 0.6315) - (2 + 0.6315)) = 1.2643 \]\[ l_4 = 0.1 \times (6 + 0.6315) = 0.66315 \] First update: \[ x(0.1) = 6 + \frac{1}{6}(1 + 2 \times 1.12 + 2 \times 1.141 + 1.264) \approx 7.056 \] \[ y(0.1) = 2 + \frac{1}{6}(0.6 + 2 \times 0.63 + 2 \times 0.6315 + 0.66315) \approx 2.632 \] Continue with step\( t = 0.1 \rightarrow t = 0.2\), using \( x = 7.056 \) and \( y = 2.632 \): Repeat calculation of \( k \)'s and \( l \)'s to derive \( x(0.2) \) and \( y(0.2) \). Final results: \[ x(0.2) \approx 8.222 \] \[ y(0.2) \approx 3.296 \]

Step 4: Graph the Solution Around t = 0

Using a numerical solver or graphing software, plot the solutions for \( x(t) \) and \( y(t) \) around \( t = 0 \) up to \( t = 0.2 \) with step length \( h = 0.1 \) for more precise visualization.

Key Concepts

Numerical Approximation

Numerical approximation is a powerful tool used to find answers when analytical solutions are hard to obtain. Instead of finding an exact result through algebraic computation, we use numerical techniques to get a close approximation. This approach is particularly important when solving differential equations, which often can't be solved exactly.

With numerical approximation:

• We approximate solutions to mathematical problems using numerical methods.
• We get results that are very close to the actual solution.
• The accuracy can be improved by refining methods or using smaller steps.

The Runge-Kutta method, a popular numerical technique, improves on the simple Euler's method by calculating several approximations at different points. This method allows for greater accuracy in solving ordinary differential equations without requiring small steps, making it much more efficient for practical applications.

Differential Equations

Differential equations play a crucial role in representing real-world phenomena where changes depend on multiple variables. Simply put, they are equations that involve an unknown function and its derivatives. In many cases, these unknown functions represent physical quantities, and the derivatives describe how these quantities change over time.Some common features include:

• They are used to model change and dynamic systems, like in physics and engineering.
• They come in two basic types: ordinary differential equations (ODEs) and partial differential equations (PDEs).
• Solving them often requires specific methods, as they can be complex.

In the context of the Runge-Kutta method, we deal with ordinary differential equations, defined by relations such as \( x'(t) = 2x - y \) and \( y'(t) = x \). These equations describe how the functions \( x(t) \) and \( y(t) \) evolve with respect to the variable \( t \). The method provides a way to estimate these changes by computing increasingly accurate approximations step-by-step.

Initial Value Problem

An initial value problem is a specific type of differential equation along with a set of initial conditions. This setup allows for the determination of a unique solution over a given interval. The initial conditions specify the state of the system at the beginning of the interval, usually denoted at \( t = 0 \), providing a starting point for the numerical approximation.Key aspects of initial value problems include:

• They consist of a differential equation and initial conditions, such as \( x(0) = 6 \) and \( y(0) = 2 \) in the given exercise.
• The solutions are determined over an interval starting from the initial condition.
• Numerical methods like Runge-Kutta are often employed to solve them.

In such problems, we compute how a system behaves starting from a known state. By incrementing small time steps (such as using \( h = 0.1 \) or \( h = 0.2 \)), the solution evolves based on the specified initial values, ensuring that each subsequent approximation builds upon the previous results.