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A hand-held calculator will suffice for Problems 1 through 10, where an initial value problem and its exact solution are given. Apply the Runge-Kutta method to approximate this solution on the interval \([0,0.5]\) with step size \(h=0.25 .\) Construct a table showing five-decimal-place values of the approximate solution and actual solution at the points \(x=0.25\) and \(0.5\). $$ y^{\prime}=-3 x^{2} y, y(0)=3 ; y(x)=3 e^{-x^{3}} $$

Short Answer

Expert verified
The approximate solutions at \(x=0.25\) and \(x=0.5\) are 2.98594 and 2.83266; exact solutions are 2.95303 and 2.64249, respectively.

Step by step solution

01

Initial Setup

We will use the Runge-Kutta method with step size \(h=0.25\) on the interval \([0, 0.5]\). The given differential equation is \(y' = -3x^2y\) with an initial condition \(y(0) = 3\). The exact solution is \(y(x) = 3e^{-x^3}\). We will find the approximate values using Runge-Kutta and compare them with the exact solution at \(x = 0.25\) and \(x = 0.5\).
02

Runge-Kutta Method at x=0.25

First, calculate the following for the first step of the method:- \(k_1 = h f(x_0, y_0) = 0.25(-3 \cdot 0^2 \cdot 3) = 0\)- \(k_2 = h f(x_0 + \frac{h}{2}, y_0 + \frac{k_1}{2}) = 0.25(-3(0.125)^2(3)) = -0.0140625\)- \(k_3 = h f(x_0 + \frac{h}{2}, y_0 + \frac{k_2}{2}) = 0.25(-3(0.125)^2(3 - 0.00703125)) = -0.014061328125\)- \(k_4 = h f(x_0 + h, y_0 + k_3) = 0.25(-3(0.25)^2(3 - 0.014061328125)) = -0.056228515625\)Now, compute the Runge-Kutta approximation for \(y(0.25)\):\[ y(0.25) \approx 3 + \frac{1}{6}(0 - 2 \cdot 0.0140625 - 2 \cdot 0.014061328125 - 0.056228515625) = 2.985938125 \]
03

Exact Solution at x=0.25

For the exact solution at \(x = 0.25\), calculate:\[ y(0.25) = 3e^{-(0.25)^3} = 3e^{-0.015625} \approx 2.953032 \]
04

Runge-Kutta Method at x=0.5

Repeat the process for the second step:- New initial values: \(x_1 = 0.25\), \(y_1 = 2.985938125\).- \(k_1 = 0.25(-3(0.25)^2(2.985938125)) = -0.0557165203125\)- \(k_2 = 0.25(-3(0.375)^2(2.985938125 - 0.02785826)) = -0.141811445\)- \(k_3 = 0.25(-3(0.375)^2(2.944009895)) = -0.139974348\)- \(k_4 = 0.25(-3(0.5)^2(2.985938125) = -0.186620725\)Compute the second approximation:\[ y(0.5) \approx 2.985938125 + \frac{1}{6}(-0.0557165203125 - 2 \cdot 0.141811445 - 2 \cdot 0.139974348 - 0.186620725) = 2.832659 \]
05

Exact Solution at x=0.5

For the exact solution at \(x = 0.5\), calculate:\[ y(0.5) = 3e^{-(0.5)^3} = 3e^{-0.125} \approx 2.642487 \]
06

Table of Results

| \(x\) | Approximate Solution | Exact Solution ||-------|------------------------|----------------|| 0.25 | 2.98594 | 2.95303 || 0.5 | 2.83266 | 2.64249 |

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Initial Value Problem
The initial value problem in differential equations involves finding a solution that satisfies a specific condition at the beginning of the interval. For the problem we're considering, the differential equation is \( y' = -3x^2y \) with the initial condition \( y(0) = 3 \). The goal is to determine the function \( y(x) \) that not only solves the differential equation but also meets the initial value provided.
This initial value gives a starting point that is crucial for ensuring the uniqueness and existence of the solution. In many practical applications, initial values can represent the state of a system at time zero, like population sizes or velocities.
Approximate Solution
An approximate solution is crucial when the exact solution to a differential equation is either very difficult or impossible to find manually. In this context, the Runge-Kutta method provides a powerful technique for numerically solving differential equations.
By utilizing a small step size, such as the \( h = 0.25 \) used here, you can generate a sequence of estimated values that approach the true path of the solution on the interval of interest, which is over \([0, 0.5]\) in this instance. The aim is to achieve values that closely mirror the true solution at specific points, like \( x = 0.25 \) and \( x = 0.5 \). This method is highly valued for its accuracy and relative efficiency.
Exact Solution
The exact solution provides a precise analytical expression for the solution of a differential equation. In this example, the exact solution is given as \( y(x) = 3e^{-x^3} \). Unlike numerical approximations, exact solutions offer a complete characterization of the system’s behavior.
Calculating this for specific values of \( x \) gives very accurate results, shown by comparing approximated values with these exact solutions at points \( x = 0.25 \) and \( x = 0.5 \). However, exact solutions can sometimes be challenging to obtain, making approximate methods like Runge-Kutta an essential tool in numerical analysis.
Differential Equation
A differential equation involves expressions that contain derivatives, defining a relationship between a function and its rates of change. Differential equations serve as fundamental tools in modeling phenomena in physics, engineering, and other sciences.
They appear in various forms, such as ordinary and partial differential equations. The given equation, \( y' = -3x^2y \), is an ordinary differential equation (ODE) because it involves ordinary derivatives. Solving this ODE provides insight into the dynamics formulated by the equation which, when partnered with initial values, helps predict future states of a system.
Numerical Analysis
Numerical analysis comprises the study and development of methods that approximate solutions to mathematical problems. It becomes indispensable when exact solutions are impractical, as in many nonlinear differential equations.
The Runge-Kutta method, which divides the problem interval into smaller steps, is a staple in numerical analysis, particularly due to its balance of simplicity and precision. By computing at intermediate points (like halfway steps), this technique refines the accuracy of the approximation, making it ideal for a wide range of applications.
Ultimately, numerical analysis helps bridge the gap between mathematical theory and real-world problem-solving with tools that deliver viable solutions in various fields.

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