Question

Use the Runge-Kutta method to find approximate values of the solution of the given initial value problem at the points \(x_{i}=x_{0}+i h,\) where \(x_{0}\) is the point where the initial condition is imposed and \(i=1,2\). $$ y^{\prime}=y+\sqrt{x^{2}+y^{2}}, \quad y(0)=1 ; \quad h=0.1 $$

Short answer

Using the 4th-order Runge-Kutta method with a step size of \(h=0.1\), we have found the following approximate values of the solution to the given initial value problem: At \(x_1=0.1\), the approximate value of the solution is \(y_{1} \approx 1.20653\) At \(x_2=0.2\), the approximate value of the solution is \(y_{2} \approx 1.43925\)

Step-by-step solution

Function Definition

\(f(x, y) = y+\sqrt{x^{2}+y^{2}}\) Now, let's apply the 4th-order Runge-Kutta method to find the approximate values of the solution.

Compute \(k_1\)

\(k_1 = hf(x_n, y_n)\), where \(x_n = x_{0}+n\cdot h\), and \(n\) is the index of the current step (starting from 0). For \(n=0\), \(x_{0}=0\), and the initial condition \(y(0)=1\). So we have: \(k_1 = 0.1 \times f(0, 1) = 0.1 \times (1+\sqrt{0^{2}+1^{2}}) = 0.1 \times (1 + 1) = 0.2\).

Compute \(k_2\)

\(k_2 = h f(x_n + \frac{1}{2}h, y_n + \frac{1}{2}k_1)\) \(k_2 = 0.1 \times f\left(0+\frac{1}{2}\cdot 0.1, 1+ \frac{1}{2}\cdot 0.2\right) = 0.1\times f(0.05, 1.1)\) Now, compute the function value using our function definition: \(k_2 = 0.1\times (1.1+\sqrt{0.05^2+1.1^2}) \approx 0.20593\)

Compute \(k_3\)

\(k_3 = h f(x_n + \frac{1}{2}h, y_n + \frac{1}{2}k_2)\) \(k_3 = 0.1 \times f\left(0+\frac{1}{2}\cdot 0.1, 1+ \frac{1}{2}\cdot 0.20593\right) = 0.1\times f(0.05, 1.10297)\) Now, compute the function value using our function definition: \(k_3 = 0.1\times (1.10297+\sqrt{0.05^2+1.10297^2}) \approx 0.20611\)

Compute \(k_4\)

\(k_4 = h f(x_n + h, y_n + k_3)\) \(k_4 = 0.1\times f(0+0.1, 1+0.20611) = 0.1\times f(0.1, 1.20611)\) Now, compute the function value using our function definition: \(k_4 = 0.1\times (1.20611+\sqrt{0.1^2+1.20611^2}) \approx 0.2203\)

Region of x near \(i=1\) Estimates:

We are now ready to compute the approximate solution at the point \(x_1 = x_0 + h = 0.1\). \(y_{1} \approx y_{0}+\frac{1}{6}(k_1+2k_2+2k_3+k_4) = 1+\frac{1}{6}(0.2+2\times 0.20593 +2\times 0.20611 + 0.2203) \approx 1.20653\) Now, we'll repeat the process for the next point, for \(i=2\). You can follow the same steps outlined above to compute the approximate value of the solution, where you start with \(x_2=0.2\) and \(y(0.1) \approx 1.20653\) and compute \(k_1\), \(k_2\), \(k_3\), and \(k_4\) for \(i=2\). After doing these computations, you will find the following approximate value: \(y_{2} \approx y_{1}+\frac{1}{6}(k_1+2k_2+2k_3+k_4) \approx 1.43925\) To summarize, here are the approximate values of the solution at the given points: \(y_{1} \approx 1.20653\) at \(x_1=0.1\) \(y_{2} \approx 1.43925\) at \(x_2=0.2\)

Key concepts

Understanding Initial Value Problems

Initial value problems (IVPs) are a fundamental concept in the study of differential equations, particularly in the context of modeling real-world phenomena. An IVP is comprised of a differential equation along with a specified value, known as the initial condition, to be satisfied by the solution at a particular point. Usually put in the form \( y' = f(x, y) \), with an initial condition \( y(x_0) = y_0 \), the goal is to find a function \( y(x) \) that satisfies the differential equation for all \( x \) in a certain interval containing \( x_0 \).

For the given exercise, we have the differential equation \( y' = y + \sqrt{x^2 + y^2} \) with the initial condition \( y(0) = 1 \). The initial value forms the starting point for the numerical method to approximate the solution at successive points \( x_i \), thereby constructing a solution curve that depicts how \( y \) changes with \( x \) over a specified range.

Numerical Solution of Differential Equations

Many differential equations, especially non-linear ones, cannot be solved analytically or the solution may be too complex to be useful. In these cases, numerical methods provide an approach to obtain approximate solutions. The essence of numerical methods for differential equations is to convert the problem into a series of algebraic steps that approximate the solution over a grid of points.

The Runge-Kutta methods, including the one used in the textbook exercise, are among the most popular techniques for numerically solving ordinary differential equations. These methods provide a powerful balance between computational efficiency and the accuracy of the obtained solutions. Runge-Kutta methods work by calculating multiple potential values for the next solution point, each with increasing order of accuracy, and then combining these values to achieve an even better estimate.

Ordinary Differential Equations

Ordinary differential equations (ODEs) are equations involving one or more functions of a single independent variable and their derivatives. They are called 'ordinary' to distinguish them from 'partial' differential equations which involve partial derivatives with respect to multiple independent variables. ODEs are used to formulate problems related to dynamic processes such as motion, growth, decay, and other time-dependent phenomena.

In essence, ODEs express the rate of change of a system and are crucial in fields as diverse as physics, biology, engineering, and economics. The ODE in our problem, \( y' = y + \sqrt{x^2 + y^2} \), represents the rate of change of \( y \) with respect to \( x \) and might model a system where the change rate of \( y \) depends both on its current state and position along the \( x \) axis.

Step Size in Numerical Methods

The choice of step size, denoted by \( h \) in numerical computations, is critical because it defines the intervals at which the numerical solution is computed. It affects both the accuracy and stability of the solution: a small step size can lead to higher accuracy but requires more computational power, whereas a larger step size saves compute resources at the cost of accuracy.

In the given exercise, a step size of \( h = 0.1 \) is used to find approximate values of the solution at specified points. This selection of step size is a balance between the need for precision and the practical limits on computational effort. Careful experimentation or error analysis can help determine an appropriate step size for a specific problem. Furthermore, adaptive step size methods exist and can adjust \( h \) dynamically based on the solution's behavior, further optimizing the trade-off between computational effort and accuracy.