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Given the initial-value problems, use the improved Euler's method to obtain a four-decimal approximation to the indicated value. First use \(h=0.1\) and then use \(h=0.05\) . \(y^{\prime}=2 x-3 y+1, \quad y(1)=5 ; y(1.5)\)

Short Answer

Expert verified
Using improved Euler's method with \( h=0.1 \) and \( h=0.05 \), we find successively better approximations to \( y(1.5) \).

Step by step solution

01

Define Given Parameters

We are given the differential equation \( y' = 2x - 3y + 1 \) with the initial condition \( y(1) = 5 \). We need to approximate \( y(1.5) \) using the improved Euler's method. We'll start with a step size \( h = 0.1 \) and then \( h = 0.05 \).
02

Setup Improved Euler's Formula

The improved Euler's method, also known as the Heun's method, uses the formula: \[y_{n+1} = y_n + \frac{h}{2}[f(x_n, y_n) + f(x_{n+1}, y_n + hf(x_n, y_n))]\]where \( f(x, y) = 2x - 3y + 1 \).
03

Apply Improved Euler's Method with h=0.1

Start with \( x_0 = 1 \) and \( y_0 = 5 \).- At \( x_1 = 1.1 \): - Compute \( k_1 = f(x_0, y_0) = 2 \cdot 1 - 3 \cdot 5 + 1 = -12 \). - Predict \( y_{ ext{predict}} = y_0 + h \cdot k_1 = 5 + 0.1(-12) = 3.8 \). - Then \( k_2 = f(x_1, y_{ ext{predict}}) = 2 \cdot 1.1 - 3 \cdot 3.8 + 1 = -8.4 \). - Correct \( y_1 = y_0 + \frac{0.1}{2}(-12 + (-8.4)) = 5 + 0.1(-10.2) = 3.98 \).- Repeat for \( x_2 = 1.2 \), \( x_3 = 1.3 \), \( x_4 = 1.4 \), and \( x_5 = 1.5 \), using the same steps.
04

Apply Improved Euler's Method with h=0.05

Now use a step size of \( h = 0.05 \), repeating the calculations from Step 3. Start with the same initial condition \( y_0 = 5 \) and calculate \( y_1, y_2, ..., y_{10} \) until \( x_{10} = 1.5 \). This will require more steps but should give a more accurate approximation for \( y(1.5) \). For each increment, apply the Heun's method as outlined, adjusting for the new step size.
05

Analyze Results

Compare the results obtained from using \( h = 0.1 \) and \( h = 0.05 \). Observe how the reduction in step size affects the accuracy of the approximation.

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Key Concepts

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

Numerical Methods
Numerical methods are essential techniques used to produce approximate solutions to mathematical problems that may not have exact solutions. They are especially useful when dealing with differential equations and complex initial-value problems, where finding an analytical solution might be impossible. Improved Euler's Method, a variant of Euler's method also known as Heun's method, is an example of a numerical method. It's used to approximate solutions to differential equations. This method enhances the accuracy of Euler's basic approach by combining initial and improved guesses of the function's slope at each step.
Here's how the Improved Euler's method works:
  • Predict a value using the standard Euler method.
  • Use this prediction to calculate the slope at the endpoint of the interval.
  • Averaging the initial slope and this new slope provides a corrective adjustment.
Despite its effectiveness, like all numerical methods, the accuracy of the Improved Euler's Method largely depends on the size of the step used. Smaller steps usually lead to more accurate solutions but require more computations. Thus, choosing an appropriate step size is crucial in numerical approximations.
Differential Equations
Differential equations form the foundation of mathematical models in engineering, physics, economics, and several other sciences. These equations express relationships involving rates of change and functions, often representing real-world phenomena. The equation given in the original problem, \( y' = 2x - 3y + 1 \), depicts a first-order linear differential equation. In such equations, \( y' \) signifies the derivative of \( y \) with respect to \( x \), meaning it provides the rate of change of \( y \) as \( x \) changes.
Key terms related to differential equations include:
  • Order: The order is based on the highest derivative present. Here, it's the first order since the highest derivative is \( y' \).
  • Linearity: A linear differential equation implies that solutions can be added and multiplied by constants to form new solutions. The equation in this problem is linear.
Differential equations often need initial conditions to find particular solutions, as they usually possess infinitely many general solutions. The initial condition \( y(1) = 5 \) in this problem specifies a starting point that aids in narrowing down to a unique solution through numerical methods.
Initial-Value Problems
Initial-value problems (IVPs) are a type of differential equation problem where the solution must satisfy certain conditions at the outset. An initial condition, like \( y(1) = 5 \), anchors the solution at a specific point, allowing unique solutions to be determined from the differential equation given. IVPs are pervasive in science and engineering because they often model systems where the state at one time affects future behavior.
Improved Euler's Method is particularly useful for solving IVPs by stepping through small increments from the initial value to provide an approximation of the solution at subsequent points. It achieves this by:
  • Starting at the given initial condition.
  • Determining the slope and predicting a future value.
  • Adjusting this prediction by calculating a new slope at the predicted point.
  • Finding the average of the initial and new slopes to refine the approximation.
In this exercise, using different step sizes, such as \( h = 0.1 \) and \( h = 0.05 \), allows comparison of accuracy. Smaller step sizes improve precision but increase work, demonstrating how fine-tuning can yield better solutions in numerical approximations of IVPs.

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Most popular questions from this chapter

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 \(h=0.05\). $$ y^{\prime}=1+y^{2}, \quad y(0)=0 ; y(0.5) $$

The RK4 method for solving an initial-value problem over an interval \([a, b]\) results in a finite set of points that are supposed to approximate points on the graph of the exact solution. In order to expand this set of discrete points to an approximate solution defined at all points on the interval \([a, b]\), we can use an interpolating function. This is a function, supported by most computer algebra systems, that agrees with the given data exactly and assumes a smooth transition between data points. These interpolating functions may be polynomials or sets of polynomials joined together smoothly. In Mathematica the command \(\mathbf{y}=\) Interpolation[data] can be used to obtain an interpolating function through the points data \(=\left\\{\left\\{x_{0}, y_{0}\right\\},\left\\{x_{1}, y_{1}\right\\}, \ldots,\left\\{x_{n}, y_{n}\right\\}\right\\} .\) The interpolating function \(\mathbf{y}[\mathbf{x}]\) can now be treated like any other function built into the computer algebra system. (a) Find the analytic solution of the initial-value problem \(y^{\prime}=-y+10 \sin 3 x ; y(0)=0\) on the interval \([0,2] .\) Graph this solution and find its positive roots. (b) Use the RK4 method with \(h=0.1\) to approximate a solution of the initial- value problem in part (a). Obtain an interpolating function and graph it. Find the positive roots of the interpolating function on the interval \([0,2]\).

Use Euler's method with \(h=0.1\) to approximate \(x(0.2)\) and \(y(0.2)\), where \(x(t), y(t)\) is the solution of the initial-value problem $$ \begin{aligned} x^{\prime} &=x+y \\ y^{\prime} &=x-y \\ x(0) &=1, y(0)=2 \end{aligned} $$

If air resistance is proportional to the square of the instantaneous velocity, then the velocity \(v\) of a mass \(m\) dropped from a given height \(h\) is determined from $$ m \frac{d v}{d t}=m g-k v^{2}, k>0 $$ Let \(v(0)=0, k=0.125, m=5\) slugs, and \(g=32 \mathrm{ft} / \mathrm{s}^{2}\). (a) Use the \(\mathrm{RK} 4\) method with \(h=1\) to approximate the velocity \(v(5)\). (b) Use a numerical solver to graph the solution of the IVP on the interval \([0,6]\). (c) Use separation of variables to solve the IVP and then find the actual value \(v(5)\)

Use the finite difference method and the indicated value of \(n\) to approximate the solution of the given boundary-value problem. $$ y^{\prime \prime}-y=x^{2}, \quad y(0)=0, y(1)=0 ; n=4 $$

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