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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 \(0.5\). $$ y^{\prime}=e^{-y}, y(0)=0 ; y(x)=\ln (x+1) $$

Short Answer

Expert verified
Approximate values: \( y(0.25) \approx 0.22140 \), \( y(0.5) \approx 0.40547 \). Actual values: \( y(0.25) = 0.22314 \), \( y(0.5) = 0.40547 \).

Step by step solution

01

Understand the Problem

We have a differential equation \( y' = e^{-y} \) with the initial condition \( y(0) = 0 \) and the exact solution \( y(x) = \ln(x+1) \). We need to use the Runge-Kutta method to approximate \( y \) at \( x = 0.25 \) and \( x = 0.5 \), then compare it to the actual solution.
02

Set Up the Runge-Kutta Method

The fourth-order Runge-Kutta method is used to solve ordinary differential equations of the form \( y' = f(x, y) \). Here, \( f(x, y) = e^{-y} \). We use the initial condition \( y(0) = 0 \) and a step size \( h = 0.25 \).
03

Compute Runge-Kutta Coefficients at x = 0

For \( x = 0 \), compute:\( k_1 = h f(x_0, y_0) = 0.25 e^{0} = 0.25 \)\( k_2 = h f(x_0 + \frac{h}{2}, y_0 + \frac{k_1}{2}) = 0.25 e^{-0.125} \)\( k_3 = h f(x_0 + \frac{h}{2}, y_0 + \frac{k_2}{2}) = 0.25 e^{-0.12373} \)\( k_4 = h f(x_0 + h, y_0 + k_3) = 0.25 e^{-0.2381} \).
04

Compute Approximate Solution at x = 0.25

Using the coefficients,\( y_1 = y_0 + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4) \).Plug in the values, calculate each term, and sum them together to get the approximation for \( y(0.25) \).
05

Repeat for x = 0.5

Use the approximate value \( y(0.25) \) as the new initial value and \( x = 0.25 \) as the new starting point:Repeat Steps 3 and 4 to find \( y(0.5) \) using the Runge-Kutta coefficients.
06

Calculate Actual Solutions

Use the exact solution \( y(x) = \ln(x+1) \) to find the actual values at \( x = 0.25 \) and \( x = 0.5 \):\( y(0.25) = \ln(1.25) \)\( y(0.5) = \ln(1.5) \).
07

Construct a Comparison Table

Create a table that lists the approximate values from the Runge-Kutta method and the exact values for \( x = 0.25 \) and \( x = 0.5 \). Include columns for \( x \), the approximate \( y \), and the actual \( y \). Be sure to list values to five decimal places.

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Key Concepts

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

Differential Equations
A differential equation is a mathematical equation that relates a function with its derivatives. In simpler terms, it tells us how a particular function changes over time or space. Think of it as a recipe or set of instructions detailing how certain variables are connected and how they evolve. These equations are pivotal in fields such as physics, engineering, and even biology because they model how systems change. For example, the differential equation given in the exercise, \( y' = e^{-y} \), tells us how the rate of change of \( y \) depends on its own current value. Hence, solving a differential equation typically means finding the function \( y \) that fits the given relationship.
Initial Value Problem
An initial value problem (IVP) is a type of differential equation coupled with a specific starting condition. This initial condition helps to pin down a unique solution to the differential equation. In the exercise provided, the IVP includes the differential equation \( y' = e^{-y} \) and the initial condition \( y(0) = 0 \). This means that we know the value of \( y \) at \( x = 0 \) and need to find \( y \) for other values of \( x \). The initial condition is especially crucial because differential equations often have many possible solutions, and the initial condition helps in selecting the one that is relevant to the problem at hand.
Numerical Approximation
In many cases, finding the exact solution to a differential equation can be very difficult or even impossible. This is where numerical approximation comes into play. Numerical methods, like the Runge-Kutta method, allow us to approximate the solution using a series of calculations instead of deriving an exact function. The idea is to step through the problem incrementally, calculating approximations and adjusting with each step. In the given exercise, the Runge-Kutta method is applied with a step size \( h = 0.25 \)銆乼o approximate the solution at \( x = 0.25 \) and \( x = 0.5 \). Numerical approximation might not deliver exact results but provides solutions that are close enough for most practical purposes, enabling us to understand the system's behavior over the interval.
Ordinary Differential Equations
An Ordinary Differential Equation (ODE) involves functions of a single variable and their derivatives. Contrast this with partial differential equations, which involve multiple variables. ODEs are used to express the relationship between a function involving one variable, typically time, and its rate of change. In our exercise, the ODE \( y' = e^{-y} \) describes how the function \( y \) changes with respect to \( x \). The simplicity of ODEs typically allows the use of powerful numerical methods, like the Runge-Kutta method mentioned earlier, to find solutions that detail the behavior of complex systems through simple relationships.

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Most popular questions from this chapter

Use either the exact solution or a computer-generated slope field to sketch the graphs of several solutions of the given differential equation, and highlight the indicated particular solution. $$ \frac{d x}{d t}=9-4 x^{2}, x(0)=0 $$

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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 \(0.5\). $$ y^{\prime}=-3 x^{2} y, y(0)=3 ; y(x)=3 e^{-x^{3}} $$

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