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 \(i=1,2\). $$ y^{\prime}+3 y=x^{2}-3 x y+y^{2}, \quad y(0)=2 ; \quad h=0.05 $$

Short answer

Answer: The approximate values of the solution at \(x_1 = 0.05\) and \(x_2 = 0.1\) using the Runge-Kutta method are \(y_1 \approx 1.646450885\) and \(y_2 \approx 1.219315453\), respectively.

Step-by-step solution

Define the function for the differential equation

In our case, the given differential equation is \(y^{\prime} + 3y = x^{2} - 3xy + y^{2}\). So our function will be: $$ F(x, y) = y^{\prime} = x^{2} - 3xy + y^{2} - 3y $$

Define variables and the step size

Now, we need to define variables and the step size: $$ x_0 = 0, \quad y_0 = 2, \quad h = 0.05 $$

Implement Runge-Kutta method of order 4 (RK4) formulas

The RK4 formulas are given by: $$ \begin{aligned} k_1 &= F(x_i, y_i), \\ k_2 &= F(x_i + 0.5h, y_i + 0.5hk_1), \\ k_3 &= F(x_i + 0.5h, y_i + 0.5hk_2), \\ k_4 &= F(x_i + h, y_i + hk_3), \\ y_{i+1} &= y_i + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4)h \end{aligned} $$ Now we need to apply the above formulas to find the values of \(y_1\) and \(y_2\).

Calculate \(y_1\)

First, we will calculate \(y_1\). Use the given function, initial values, and RK4 formulas: $$ \begin{aligned} k_1 &= F(0, 2) = 0^2 - 3(0)(2) + 2^2 - 3(2) = -6, \\ k_2 &= F(0+0.5(0.05), 2 + 0.5(0.05)(-6)) = F(0.025, 1.85) = -7.19375, \\ k_3 &= F(0.025, 2 - 0.5(0.05)(7.19375)) = F(0.025, 1.8209375) = -7.471601563, \\ k_4 &= F(0.05, 2 - 0.05(7.471601563)) = F(0.05, 1.625919923) = -9.084614625, \\ y_{1} &= 2 + \frac{1}{6}((-6) + 2(-7.19375) + 2(-7.471601563) + (-9.084614625))(0.05) = 1.646450885 \end{aligned} $$ Thus, the approximate value of the solution at \(x_1 = 0.05\) is \(y_1 \approx 1.646450885\).

Calculate \(y_2\)

Now, we will calculate \(y_2\), using the obtained \(y_1\) value and the RK4 formulas: $$ \begin{aligned} k_1 &= F(0.05, 1.646450885) = -8.952203227, \\ k_2 &= F(0.075, 1.646450885 + 0.5(0.05)(-8.952203227)) = F(0.075, 1.394304303) = -11.46268665, \\ k_3 &= F(0.075, 1.646450885 - 0.5(0.05)(11.46268665)) = F(0.075, 1.357034112) = -11.90356922, \\ k_4 &= F(0.1, 1.646450885 - 0.05(11.90356922)) = F(0.1, 1.047224527) = -14.78753295, \\ y_{2} &= 1.646450885 + \frac{1}{6}((-8.952203227) + 2(-11.46268665) + 2(-11.90356922) + (-14.78753295))(0.05) = 1.219315453 \end{aligned} $$ Thus, the approximate value of the solution at \(x_2 = 0.1\) is \(y_2 \approx 1.219315453\). In conclusion, the approximate values of the solution at \(x_1 = 0.05\) and \(x_2 = 0.1\) using the Runge-Kutta method are \(y_1 \approx 1.646450885\) and \(y_2 \approx 1.219315453\), respectively.

Key concepts

Initial Value Problem

When studying the concept of an initial value problem (IVP), we are essentially delving into a fundamental scenario within differential equations. An IVP is characterized by a differential equation coupled with a specified value, known as the initial condition, at a particular point. This starting point is critical, as it serves as the foundation for determining the solution of the differential equation.

The scenario provided in our exercise is a prime example of an initial value problem, where the differential equation is given by
\[ y' + 3y = x^2 - 3xy + y^2, \]and the initial condition is
\[ y(0) = 2. \]Here, the solution must not only satisfy the differential equation, but also align with the initial condition at
\(x_0 = 0\). This constraint helps in finding a unique solution, a critical aspect since differential equations can have infinitely many solutions without such a condition. The IVP is key in various applications, such as physics and engineering, where initial states often dictate subsequent behavior.

Differential Equations

Differential equations form the backbone of modeling changes over continuous domains. They are dynamic mathematical entities that relate functions to their derivatives, essentially expressing how rates of change influence the behavior of various systems. In our Runge-Kutta exercise, the differential equation is nonlinear and somewhat complex:
\(y' + 3y = x^2 - 3xy + y^2 \).

Understanding differential equations involves discerning both the type (ordinary, partial, etc.) and order (first, second, etc.). The primary aim is to find the function
\(y(x)\) that satisfies the equation across a domain of interest. In our case, it's the 'ordinary differential equation' because it contains one independent variable, which is
\(x\).

Integral to fields like physics, engineering, economics, and biology, differential equations equip us with the means to predict and explain real-world phenomena, such as motion, growth, decay, and oscillations, mirroring the complexity inherent in the systems they describe.

Numerical Methods

Numerical methods serve as powerful tools to solve mathematical problems that may be intractable by analytical means, especially concerning differential equations. These methods approximate solutions to complex problems through a computational lens.

In the context of our exercise, the Runge-Kutta method is a numerical technique used to approximate the solution to initial value problems involving ordinary differential equations. Its attraction lies in its balance between accuracy and computational tractability. The fourth-order Runge-Kutta method (RK4), which we utilized in the provided solution, is lauded for its precision and efficiency, enabling us to estimate solutions at specific points, like
\(y_1\) and
\(y_2\), with notable accuracy.

Key Aspects of Numerical Methods:

• They involve discretizing the continuous domain and performing calculations at selected points.
• They offer iterative or formula-based approaches to systematically improve upon initial approximations.
• They can deal with various types of equations and constraints, including those stemming from boundary conditions.

This discipline is crucial today as it provides actionable solutions to real-life problems where direct analytical solutions may not be possible or realistic.