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 \(i=1,2\). $$ y^{\prime}=\frac{1+x}{1-y^{2}}, \quad y(2)=3 ; \quad h=0.1 $$

Short answer

Question: Using the Runge-Kutta method, find the approximate values of the solution to the following initial value problem: \(y^{\prime}=\frac{1+x}{1-y^{2}}\), \(y(2)=3\), at the points \(x_{1}=2.1\) and \(x_{2}=2.2\), with step size \(h=0.1\). Answer: The approximate values of the solution at the points \(x_{1}=2.1\) and \(x_{2}=2.2\) are \(y_{1}\approx3.002823\) and \(y_{2}\approx3.005626\), respectively.

Step-by-step solution

Identify the differential equation and initial condition

The given differential equation is \(y^{\prime}=\frac{1+x}{1-y^{2}}\) with initial condition \(y(2)=3\). Our task is to find the values of the solution at \(x_{1}=x_{0}+0.1\) and \(x_{2}=x_{0}+0.2\), where \(x_{0}=2\).

Write down the Runge-Kutta method formula

The formula for the Runge-Kutta method of order 4 (which is most commonly used) involves calculating four slopes, k1, k2, k3, and k4, and updating the function value y as follows: $$ k_{1} = h f(x_{i}, y_{i}), $$ $$ k_{2} = h f(x_{i} + \frac{h}{2}, y_{i} + \frac{k_{1}}{2}), $$ $$ k_{3} = h f(x_{i} + \frac{h}{2}, y_{i} + \frac{k_{2}}{2}), $$ $$ k_{4} = h f(x_{i} + h, y_{i} + k_{3}), $$ $$ y_{i+1} = y_{i} + \frac{1}{6} (k_{1} + 2k_{2} + 2k_{3} + k_{4}). $$

Calculate k1, k2, k3, k4 for i=1

Since we know the initial value at \(x_{0}=2\) and \(y_{0}=3\), we can calculate the k-values for i=1: - For k1: $$ k_{1} = 0.1 \cdot \frac{1+2}{1-3^{2}} = -\frac{0.1}{8} $$ - For k2: $$ k_{2} = 0.1 \cdot \frac{1 + (2+0.05)}{1-(3+\frac{-0.1}{16})^{2}} = 0.1 \cdot \frac{3.05}{1-(3-\frac{1}{16})^{2}} \approx -0.012413 $$ - For k3: $$ k_{3} = 0.1 \cdot \frac{1+(2+0.05)}{1-(3+\frac{-0.012413}{2})^{2}} = 0.1 \cdot \frac{3.05}{1-(2.99379)^{2}} \approx -0.012402 $$ - For k4: $$ k_{4} = 0.1 \cdot \frac{1+(2+0.1)}{1-(3-0.012402)^{2}} = 0.1 \cdot \frac{3.1}{1-(2.9876)^{2}} \approx -0.024622 $$

Calculate y1 using the Runge-Kutta update formula

Using the Runge-Kutta update formula, we can calculate y1: $$ y_{1} = y_{0} + \frac{1}{6} (k_{1} + 2k_{2} + 2k_{3} + k_{4}) \approx 3 - \frac{1}{6}(-\frac{1}{8}-2 \cdot 0.012413 - 2 \cdot 0.012402 - 0.024622) \approx 3.002823 $$ So the approximate value of y at \(x_{1}=2.1\) is \(y_{1} \approx 3.002823\).

Calculate k1, k2, k3, k4 for i=2

Now that we have the value of y1, we can calculate the k-values for i=2: - For k1: $$ k_{1} = 0.1 \cdot \frac{1+2.1}{1-3.002823^{2}} \approx -0.024708 $$ - For k2: $$ k_{2} = 0.1 \cdot \frac{1 + (2.1+0.05)}{1-(3.002823-\frac{-0.024708}{2})^{2}} \approx -0.037043 $$ - For k3: $$ k_{3} = 0.1 \cdot \frac{1+(2.1+0.05)}{1-(3.002823+\frac{-0.037043}{2})^{2}} \approx -0.037026 $$ - For k4: $$ k_{4} = 0.1 \cdot \frac{1+(2.1+0.1)}{1-(3.002823-0.037026)^{2}} \approx -0.048868 $$

Calculate y2 using the Runge-Kutta update formula

Using the Runge-Kutta update formula, we can calculate y2: $$ y_{2} = y_{1} + \frac{1}{6} (k_{1} + 2k_{2} + 2k_{3} + k_{4}) \approx 3.002823 - \frac{1}{6}(-0.024708 - 2 \cdot 0.037043 - 2 \cdot 0.037026 - 0.048868) \approx 3.005626 $$ So the approximate value of y at \(x_{2}=2.2\) is \(y_{2} \approx 3.005626\). Thus, the approximate values of the solution at the points \(x_{1}=2.1\) and \(x_{2}=2.2\) are \(y_{1}\approx3.002823\) and \(y_{2}\approx3.005626\), respectively.

Key concepts

Initial Value Problem

An initial value problem (IVP) in mathematics is a type of differential equation along with a specified value, called the initial condition, at a given point. Specifically, it consists of an ordinary differential equation (ODE) and the value of the unknown function at a certain point.

For example, if we have an equation describing how a quantity changes over time, the initial value problem would include that equation along with the value of the quantity at time zero. The exercise we're discussing involves the IVP \(y'(x) = \frac{1+x}{1-y^2}\) with the initial condition \(y(2)=3\). This means we know the rate of change of \(y\) at any point \(x\) and are given \(y\)'s value when \(x=2\).

Solving an IVP helps us predict the future behavior of the system described by the differential equation, based on the initial condition. The Runge-Kutta method is a powerful tool for approximating the solution to such problems, particularly when an exact solution is difficult or impossible to find.

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They are essential in mathematically modeling and understanding phenomena where quantities change in relation to one another. Differential equations can be used to describe a wide range of real-world situations such as motion, growth, decay, or changes in temperature.

In our exercise, the differential equation is \(y'(x) = \frac{1+x}{1-y^2}\). This equation tells us the rate at which \(y\) is changing at any point \(x\). The solution to a differential equation is the function \(y(x)\) itself, which, once known, gives us complete knowledge of the behavior of the system from the given point onwards. In many cases, these equations cannot be solved analytically, and that's where numerical methods like the Runge-Kutta method come into play to provide an approximate solution.

Numerical Approximation

Numerical approximation involves finding an approximate solution to a mathematical problem when an exact solution is not practicable. This method is integral to computational mathematics and is commonly used when dealing with real-world problems where data may be incomplete or the systems too complex.

With the Runge-Kutta method, we gain a numerical approximation to the solution of differential equations. It works by taking a small step from the initial value using the slope of the function (derived from the differential equation), then averaging slopes across this step to improve the estimate. In the case of the given exercise, a step size \(h=0.1\) is used to incrementally find the approximate solution at \(x_1\) and \(x_2\).

This process is repeated, with each step's calculation depending on the previous, thus progressing the approximation across the interval of interest. The exercise demonstrates this iterative process, providing a practical approach to understanding the behavior of complex systems when direct analytical solutions are unattainable.