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

Short answer

Question: Using the fourth-order Runge-Kutta method, find the approximate solutions of the initial value problem \(y' = 2x^2 + 3y^2 - 2\) with \(y(2) = 1\) and step size \(h = 0.05\) at points \(x_1 = 2.05\) and \(x_2 = 2.10\). Answer: The approximate solutions of the given initial value problem at \(x_1 = 2.05\) is \(y_1 \approx 1.0335\) and at \(x_2 = 2.10\) it is \(y_2 \approx 1.0692\).

Step-by-step solution

Define the given differential equation and initial conditions

The given differential equation is \(y'=2x^2 + 3y^2 - 2\). The initial condition is \(y(2)=1\). Let \(f(x,y)=2x^2 + 3y^2 -2\). We want to find the approximate solution at \(x_1\) and \(x_2\).

Set up the fourth-order Runge-Kutta method

In the fourth-order Runge-Kutta method, we update our estimate of the solution at each step as follows: \begin{align*} k_{1} &= h f(x_n, y_n), \\ k_{2} &= h f(x_n + \frac{1}{2}h, y_n + \frac{1}{2}k_{1}), \\ k_{3} &= h f(x_n + \frac{1}{2}h, y_n + \frac{1}{2}k_{2}), \\ k_{4} &= h f(x_n + h, y_n + k_{3}), \\ y_{n+1} &= y_n + \frac{1}{6}(k_{1} + 2k_{2} + 2k_{3} + k_{4}). \end{align*}Using this method, we will update our estimate for the solution at \(x_1 = x_0 + h = 2+0.05=2.05\) and \(x_2=x_0+2h=2+2(0.05)=2.10.\)

Calculate \(k_1, k_2, k_3, k_4,\) and \(y_1\)

First, we'll find the \(k_i\) values and the approximate solution \(y_1\) at \(x_1=2.05\). We start with the given \(x_n=2\) and \(y_n=1\):\begin{align*} k_{1} &= h f(2, 1) = 0.05(2(2)^2 + 3(1)^2 - 2) = 0.05(12) = 0.6, \\ k_{2} &= 0.05 f(2 + \frac{1}{2}(0.05), 1 + \frac{1}{2}(0.6)) = 0.05 f(2.025, 1.3) \approx 0.0531, \\ k_{3} &= 0.05 f(2 + \frac{1}{2}(0.05), 1 + \frac{1}{2}(0.0531)) \approx 0.0531, \\ k_{4} &= 0.05 f(2 + 0.05, 1 + 0.0531) \approx 0.0629, \\ y_{1} &= 1 + \frac{1}{6}(0.6 + 2(0.0531) + 2(0.0531) + 0.0629) \approx 1.0335. \end{align*}

Calculate \(k_1, k_2, k_3, k_4,\) and \(y_2\)

Now we'll find the \(k_i\) values and the approximate solution \(y_2\) at \(x_2=2.10\). We use the previous result at \(x_1=2.05\) as our input, with \(x_n=2.05\) and \(y_n \approx 1.0335\):\begin{align*} k_{1} &= h f(2.05, 1.0335) \approx 0.065, \\ k_{2} &= 0.05 f(2.05 + \frac{1}{2}(0.05), 1.0335 + \frac{1}{2}(0.065)) \approx 0.066, \\ k_{3} &= 0.05 f(2.05 + \frac{1}{2}(0.05), 1.0335 + \frac{1}{2}(0.066)) \approx 0.066, \\ k_{4} &= 0.05 f(2.05 + 0.05, 1.0335 + 0.066) \approx 0.075, \\ y_{2} &= 1.0335 + \frac{1}{6}(0.065 + 2(0.066) + 2(0.066) + 0.075) \approx 1.0692. \end{align*}

Present the final approximate solutions

The approximate solution of the given initial value problem using the fourth-order Runge-Kutta method at \(x_1 = 2.05\) is \(y_1 \approx 1.0335\) and at \(x_2 = 2.10\) it is \(y_2 \approx 1.0692\).

Key concepts

Understanding Initial Value Problems

An initial value problem (IVP) is a fundamental concept in the study of ordinary differential equations, a core area of applied mathematics that models numerous phenomena in science and engineering. An IVP consists of a differential equation along with a specific value, called the initial condition, which the solution must satisfy at a designated point, typically referred to as the initial point.

For example, in the IVP given by the differential equation y'=2x^2 + 3y^2 - 2 with an initial condition y(2)=1, the goal is to find a function y(x) that not only solves the equation but also passes through the point (2, 1) on the graph. Due to the complexity of some equations, finding an exact solution analytically can be challenging, which is where numerical methods like the Runge-Kutta method become extremely useful.

The Essence of Differential Equations

Differential equations are equations that involve an unknown function and its derivatives. They are a cornerstone in understanding how quantities change and are related to one another, capturing dynamics in various contexts like physics, biology, economics, and beyond.

A differential equation like y'=2x^2 + 3y^2 - 2 expresses a relationship between a function y(x) and its derivative y'. The equation gives us a rule for the growth rate of y depending on the current values of x and y. To solve such an equation, one would typically look for a function whose rate of change (y') at every point x follows the pattern described by the equation.

Numerical Methods: Navigating Through Approximations

Numerical methods are techniques used to obtain approximate solutions to mathematical problems that often cannot be solved analytically. They enable mathematicians and scientists to approximate answers to very precise degrees, providing practical solutions where theoretical ones are infeasible. The Runge-Kutta method is one such numerical approach, particularly renowned for solving initial value problems for ordinary differential equations.

The Runge-Kutta method we're discussing is a fourth-order method, indicating its high level of accuracy. It estimates the solution by combining several intermediate values, calculated using increments (k_i), to advance step by step from the initial point. By carefully choosing the size of each step (h), the Runge-Kutta method balances computational workload with the desired precision, making it a robust tool for tackling complex differential equations.