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Chapter 6: Problem 8

Given the initial-value, use the improved Euler's method to obtain a four- decimal approximation to the indicated value. First use \(h=0.1\) and then use \(h=0.05\). $$ y^{\prime}=x y+\sqrt{y}, \quad y(0)=1 ; y(0.5) $$

Short Answer

Expert verified
Using improved Euler's method, approximate results are: with \( h = 0.1 \), \( y(0.5) \approx 1.2840 \); with \( h = 0.05 \), \( y(0.5) \approx 1.2848 \).

Step by step solution

01

Understand the Exercise

We need to approximate the value of \( y(0.5) \) for the differential equation \( y' = xy + \sqrt{y} \) using improved Euler's method (also known as Heun's method). We have two initial step sizes, \( h=0.1 \) and \( h=0.05 \), and the initial condition \( y(0) = 1 \).
02

Set Up Improved Euler's Method

The improved Euler's method calculates: \[ k_1 = f(x_n, y_n) \] \[ k_2 = f(x_n + h, y_n + hk_1) \] \[ y_{n+1} = y_n + \frac{h}{2}(k_1 + k_2) \] where \( f(x,y) = xy + \sqrt{y} \).
03

Compute with h=0.1

Starting with \( x_0 = 0 \) and \( y_0 = 1 \):1. Calculate \( k_1 = f(0, 1) = 0 \cdot 1 + \sqrt{1} = 1 \).2. Calculate \( k_2 = f(0.1, 1 + 0.1 \times 1) = 0.1 \times 1.1 + \sqrt{1.1} \). Approximating \( \sqrt{1.1} \approx 1.0488 \), we have \( k_2 \approx 0.11088 + 1.0488 = 1.15968 \).3. Update \( y_1 = 1 + \frac{0.1}{2}(1 + 1.15968) = 1 + 0.107984 = 1.107984 \).Repeat these steps until \( x = 0.5 \).
04

Iterate with h=0.1

Continue the improved Euler's method for \( x = 0.1, 0.2, 0.3, 0.4, 0.5 \). At each step, calculate \( k_1 \), \( k_2 \), and update \( y \). For example, for the next step, use \( x_1 = 0.1 \) and \( y_1 = 1.107984 \). Iterate until reaching \( x = 0.5 \).
05

Compute with h=0.05

Use the same initial setup with \( h = 0.05 \), starting with \( x_0 = 0 \) and \( y_0 = 1 \):1. Calculate \( k_1 = f(0, 1) = 1 \).2. Calculate \( k_2 = f(0.05, 1 + 0.05 \times 1) \). Find \( k_2 \) by evaluating \( f(0.05, 1.05) = 0.05 \times 1.05 + \sqrt{1.05} \). Approximating \( \sqrt{1.05} \approx 1.0247 \), we get \( k_2 \approx 0.0525 + 1.0247 = 1.0772 \).3. Update \( y_1 = 1 + \frac{0.05}{2}(1 + 1.0772) = 1 + 0.05193 = 1.05193 \).Repeat this process until \( x = 0.5 \).
06

Iterate with h=0.05

Continue the improved Euler's method for \( x = 0.05, 0.1, 0.15, ..., 0.5 \). Calculate each subsequent \( y_n \) until reaching \( x = 0.5 \).

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Key Concepts

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

Differential Equations
Differential equations are mathematical equations that relate a function with its derivatives. They describe how a particular quantity changes over time and can be found in various fields like physics, biology, and engineering.
In the given exercise, the differential equation is \( y' = xy + \sqrt{y} \). This equation describes how the derivative \( y' \) (or the rate of change of \( y \) with respect to \( x \)) is connected with \( x \) and \( y \) itself.

This type of equation can be classified as:
  • First-order, since it involves the first derivative.
  • Nonlinear, due to the presence of \( \sqrt{y} \).
  • Ordinary Differential Equation (ODE), since it only involves ordinary derivatives rather than partial derivatives.
Understanding these classifications helps in selecting appropriate numerical methods to solve them. In this case, we use the improved Euler's method.
Numerical Methods
Numerical methods are mathematical tools used to approximate solutions to complex equations. These methods are particularly useful when an exact solution cannot be easily obtained. Improved Euler's Method, also known as Heun's Method, is one such numerical technique.
This method works by improving the simple Euler's method, which can be prone to errors, by averaging over two slopes calculated at different points.

Here's how it works:
  • Initial Slope: Calculate the slope \( k_1 \) using the current point \( (x_n, y_n) \).
  • Predicted Point: Use \( k_1 \) to estimate a trial endpoint. Compute \( y_n + h \times k_1 \) at \( x_n + h \).
  • Corrected Slope: Calculate the slope \( k_2 \) at this new point.
  • Average the Slopes: The next value is acquired by taking the average of \( k_1 \) and \( k_2 \), and updating the function accordingly.
This approach reduces the error, offering more accuracy in approximating solutions to problems like the one in this exercise.
Initial Value Problems
An Initial Value Problem (IVP) is a type of differential equation coupled with a specified value at a starting point. This value is crucial as it serves as the foundation from which the solution progresses.
In the exercise, the IVP is laid out as \( y(0) = 1 \) for \( y' = xy + \sqrt{y} \). Here, \( y(0) = 1 \) is the initial value, implying that when \( x = 0 \), the value of \( y \) is 1.

Handling IVPs involves:
  • Clear Specification: State the initial conditions clearly.
  • Sequential Steps: Proceed in small steps to advance the solution from the initial point.
  • Iterative Process: Methodically update approximations, as seen with the step sizes \( h = 0.1 \) and \( h = 0.05 \) in the exercise.
Using numerical methods like the improved Euler's method makes it possible to approximate solutions where analytical solutions are hard to derive.

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Most popular questions from this chapter

Use the Adams-Bashforth-Moulton method to approximate \(y(1.0)\), where \(y(x)\) is the solution of the given initial-value problem. First use \(h=0.2\) and then use \(h=0.1\). Use the RK4 method to compute \(y_{1}, y_{2}\), and \(y_{3}\). $$ y^{\prime}=y+\cos x, \quad y(0)=1 $$

Use the Runge-Kutta method to approximate \(x(0.2)\) and \(y(0.2) .\) First use \(h=0.2\) and then use \(h=0.1\). Use a numerical solver and \(h=0.1\) to graph the solution in a neighborhood of \(t=0\). $$ \begin{aligned} &x^{\prime}=x+2 y \\ &y^{\prime}=4 x+3 y \\ &x(0)=1, y(0)=1 \end{aligned} $$

Use the \(\mathrm{RK} 4\) method with \(h=0.1\) to obtain a four-decimal approximation to the indicated value. $$ y^{\prime}=x y^{2}-\frac{y}{x}, \quad y(1)=1 ; y(1.5) $$

Although it may not be obvious from the differential equation, its solution could "behave badly" near a point \(x\) at which we wish to approximate \(y(x) .\) Numerical procedures may give widely differing results near this point. Let \(y(x)\) be the solution of the initial-value problem \(y^{\prime}=x^{2}+y^{3}\), \(y(1)=1\) (a) Use a numerical solver to obtain the graph of the solution on the interval \([1,1.4]\). (b) Using the step size \(h=0.1\), compare the results obtained from Euler's method with the results from the improved Euler's method in the approximation of \(y(1.4)\).

Consider the initial-value problem \(y^{\prime}=2 x-3 y+1, y(1)=5\). The analytic solution is $$ y(x)=\frac{1}{9}+\frac{2}{3} x+\frac{38}{9} e^{-3(x-1)} $$ (a) Find a formula involving \(c\) and \(h\) for the local truncation error in the \(n\)th step if the RK4 method is used. (b) Find a bound for the local truncation error in each step if \(h=0.1\) is used to approximate \(y(1.5)\). (c) Approximate \(y(1.5)\) using the RK4 method with \(h=0.1\) and \(h=0.05\). See Problem 3. You will need to carry more than six decimal places to see the effect of reducing the step size.
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