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Chapter 2: Problem 7

An initial value problem and its exact solution \(y(x)\) are given. Apply Euler's method twice to approximate to this solution on the interval \(\left[0, \frac{1}{2}\right]\), first with step size \(h=0.25\), then with step size \(h=0.1 .\) Compare the threedecimal-place values of the two approximations at \(x=\frac{1}{2}\) with the value \(y\left(\frac{1}{2}\right)\) of the actual solution. $$ y^{\prime}=-3 x^{2} y, v(0)=3 ; y(x)=3 e^{-x^{3}} $$

Short Answer

Expert verified
With a smaller step size, the approximation is closer to the exact value.

Step by step solution

01

Understand the Initial Value Problem

We are given the differential equation \(y' = -3x^2y\) with the initial condition \(y(0) = 3\). The exact solution to this initial value problem is \(y(x) = 3e^{-x^3}\). Our task is to approximate and compare Euler's method results for different step sizes with the exact solution at \(x=0.5\).
02

Euler's Method Setup

Euler's method formula is given by \(y_{n+1} = y_n + h f(x_n, y_n)\), where \(f(x, y) = y' = -3x^2y\) is the differential equation. We will use this formula to approximate \(y(x)\) on \([0, 0.5]\) with step sizes \(h = 0.25\) and \(h = 0.1\).
03

Euler's Method with Step Size h=0.25

**Initial Conditions:** \(x_0 = 0, y_0 = 3\) **Step 1:** At \(x_1 = 0.25\), \(y_1 = y_0 + 0.25(-3(0)^2)(3) = 3\) **Step 2:** At \(x_2 = 0.5\), \(y_2 = y_1 + 0.25(-3(0.25)^2)(3) = 3 - 0.25(0.1875)(3) = 3 - 0.140625 = 2.859\) The approximation for \(x = 0.5\) with step size \(h = 0.25\) is \(y(0.5) \approx 2.859\).
04

Euler's Method with Step Size h=0.1

**Initial Conditions:** \(x_0 = 0, y_0 = 3\) **Step 1:** At \(x_1 = 0.1\), \(y_1 = y_0 + 0.1(-3(0)^2)(3) = 3\) **Step 2:** At \(x_2 = 0.2\), \(y_2 = y_1 + 0.1(-3(0.1)^2)(3) = 3 - 0.03 = 2.97\) **Step 3:** At \(x_3 = 0.3\), \(y_3 = y_2 + 0.1(-3(0.2)^2)(2.97) = 2.97 - 0.03564 = 2.93436\) **Step 4:** At \(x_4 = 0.4\), \(y_4 = y_3 + 0.1(-3(0.3)^2)(2.93436) = 2.93436 - 0.07953672 = 2.85482328\) **Step 5:** At \(x_5 = 0.5\), \(y_5 = y_4 + 0.1(-3(0.4)^2)(2.85482328) = 2.85482328 - 0.13619488 = 2.7186284\) The approximation for \(x = 0.5\) with step size \(h = 0.1\) is \(y(0.5) \approx 2.7186284\).
05

Calculate the Exact Solution at x=0.5

Using the exact solution \(y(x) = 3e^{-x^3}\), we compute: \(y(0.5) = 3e^{-0.125} \approx 3(0.8869) \approx 2.6607\).
06

Compare Approximations with Exact Value

With step size \(h=0.25\), \(y(0.5) \approx 2.859\). With step size \(h=0.1\), \(y(0.5) \approx 2.7186284\). Exact value \(y(0.5) \approx 2.6607\). The approximation with \(h=0.1\) is closer to the exact value than with \(h=0.25\).

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Key Concepts

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

Initial Value Problem
An Initial Value Problem (IVP) in mathematics is a differential equation coupled with a specific value, known as the initial condition, at a starting point. In this scenario, our given differential equation is \(y' = -3x^2y\), and the initial condition is \(y(0) = 3\). This means when \(x = 0\), the value of \(y\) is 3. The purpose of stating an initial value is to provide a starting point, which sets the groundwork for solving the differential equation uniquely. Given the initial condition, we can progress to find not just one, but a specific solution.
  • Initial condition: Supplies the starting value at a given point.
  • Forms the basis for applying numerical methods, like Euler's method to find a solution over an interval.
  • Ensures a unique solution for the differential equation is achieved when solving analytically.
Differential Equations
Differential equations are mathematical equations that involve functions and their derivatives. They describe a relationship between a function and its rates of change, capturing dynamic processes. In our problem, \(y' = -3x^2y\), we have a first-order differential equation where the rate of change of the function \(y\) with respect to \(x\) depends on both \(x\) and \(y\). This ties the behavior of the function \(y\) directly to its position along \(x\).There are two common ways to solve differential equations:
  • Analytically: Finding a formula or exact solution like \(y(x) = 3e^{-x^3}\) which precisely solves the equation.
  • Numerically: Using approximation methods such as Euler鈥檚 method to approximate the solution at discrete points.
Numerical Approximation
Numerical approximation involves estimating solutions to problems that may not be solvable analytically. In situations where an exact analytic solution is complicated or impossible to obtain, numerical methods such as Euler's method are employed.Euler's method provides a simple technique to approximate solutions for differential equations. By breaking down the interval of interest into small steps (known as "step size"), it estimates the function's value at successive points starting from the initial condition.Key Factors:
  • Step Size \(h\): Determines the accuracy of the approximation; smaller step sizes generally give more precise results.
  • Iteration: Repeatedly apply the method's formula \(y_{n+1} = y_n + hf(x_n, y_n)\) to calculate approximate solutions.
  • Trade-off: Smaller step sizes provide greater accuracy but increase computational effort.
Exact Solution
The exact solution of a differential equation provides the precise mathematical form of the solution without approximation, valid over a continuous range or domain. In our exercise, the exact solution is \(y(x) = 3e^{-x^3}\). This expression exactly satisfies the initial value problem and allows us to determine \(y\) for any \(x\) within the interval.Advantages of an Exact Solution:
  • Accuracy: Represents the true behavior of the system described by the differential equation.
  • Comparison: Serves as a benchmark against which numerical approximation results can be compared.
  • Insights: Provides a formulaic representation, offering deep insights into the characteristics of the solution, such as asymptotic behavior.

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Most popular questions from this chapter

Find the exact solution of the given initial value problem. Then apply Euler's method twice to approximate (to four decimal places) this solution on the given interval, first with step size \(h=0.01\), then with step size \(h=0.005 .\) Make a table showing the approximate values and the actual value, together with the percentage error in the more accurate approximation, for \(x\) an integral multiple of 0.2. Throughout, primes denote derivatives with respect to \(x\). $$ y y^{\prime}=2 x^{3}, y(1)=3 ; 1 \leqq x \leqq 2 $$

Consider a population \(P(t)\) satisfying the extinctionexplosion equation \(d P / d t=a P^{2}-b P\), where \(B=a P^{2}\) is the time rate at which births occur and \(D=b P\) is the rate at which deaths occur. If the initial population is \(P(0)=P_{0}\) and \(B_{0}\) births per month and \(D_{0}\) deaths per month are occurring at time \(t=0\), show that the threshold population is \(M=D_{0} P_{0} / B_{0}\).

A programmable calculator or a computer will be useful for Problems 11 through \(16 .\) In each problem find the exact solution of the given initial value problem. Then apply the RungeKutta method twice to approximate (to five decimal places) this solution on the given interval, first with step size \(h=0.2\), then with step size \(h=0.1 .\) Make a table showing the approximate values and the actual value, together with the percentage error in the more accurate approximation, for \(x\) an integral multiple of 0.2. Throughout, primes denote derivatives with respect to \(x\). \(y^{\prime}=\frac{1}{2}(y-1)^{2}, y(0)=2 ; 0 \leqq x \leqq 1\)

A computer with a printer is required for Problems 17 through 24\. In these initial value problems, use the Runge-Kutta method with step sizes \(h=0.2,0.1,0.05\), and \(0.025\) to approximate to six decimal places the values of the solution at five equally spaced points of the given interval. Print the results in tabular form with appropriate headings to make it easy to gauge the effect of varying the step size h. Throughout, primes denote derivatives with respect to \(x\). \(y^{\prime}=x^{2 / 3}+y^{2 / 3}, y(0)=1 ; 0 \leqq x \leqq 2\)

Apply Euler's method with successively smaller step sizes on the interval \([0,2]\) to verify empirically that the solution of the initial value problem $$ \frac{d y}{d x}=x^{2}+y^{2}, \quad y(0)=0 $$ has a vertical asymptote near \(x=2.003147 .\) (Contrast this with Example 2 , in which \(y(0)=1\).)
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