/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 The solution to the initial valu... [FREE SOLUTION] | 91Ó°ÊÓ

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The solution to the initial value problem $$ y^{\prime}=\frac{2}{x^{4}}-y^{2}, \quad y(1)=-0.414 $$ crosses the x-axis at a point in the interval 31, 24 . By experimenting with the fourth-order Runge-Kutta subroutine, determine this point to two decimal places.

Short Answer

Expert verified
By implementing the Runge-Kutta method and linear interpolation, the x-coordinate of the point where the solution y crosses the x-axis can be accurately estimated to two decimal places.

Step by step solution

01

Restate the initial value problem

The initial value problem is given by the differential equation \(y^{\prime}=\frac{2}{x^{4}}-y^{2}\) with the initial condition \(y(1)=-0.414\).
02

Implement the Runge-Kutta method

Define the function \(f(x,y) = \frac{2}{x^{4}} - y^{2}\). Create a while loop starting at x=1 with y=-0.414 and increment x in small steps. In each iteration, implement the Runge-Kutta method as follows: \n1. Calculate \( k_{1} = h \cdot f(x, y) \).2. Calculate \( k_{2} = h \cdot f(x + 0.5 \cdot h, y + 0.5 \cdot k_{1}) \).3. Calculate \( k_{3} = h \cdot f(x + 0.5 \cdot h, y + 0.5 \cdot k_{2}) \).4. Calculate \( k_{4} = h \cdot f(x + h, y + k_{3}) \).5. Update \( y = y + \frac{1}{6} \cdot (k_{1} + 2 \cdot k_{2} + 2 \cdot k_{3} + k_{4}) \).\nStop the loop once \(y\) changes from negative to positive. Then, calculate the zero crossing by linear interpolation between the two points surrounding \(y = 0\).
03

Estimate the zero-crossing

Once the change in sign of \(y\) is detected, use linear interpolation to estimate \(x\) for \(y = 0\). This is achieved by creating the line equation between the two points \((x_{0}, y_{0})\) and \((x_{1}, y_{1})\) surrounding \(y = 0\), and solving \(y = 0\) for \(x\). It results in \(x = x_{0} - y_{0} \cdot \frac{x_{1} - x_{0}}{y_{1} - y_{0}}\).

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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 is a differential equation accompanied by an initial condition. This condition specifies the value of the solution at a particular point, often denoted as the starting point. In our exercise, the initial value problem involves the differential equation \( y^{\prime}=\frac{2}{x^{4}}-y^{2} \) with the initial condition \( y(1)=-0.414 \).
  • Initial Condition: This is information about the value of the function at a particular point, which is crucial for finding a specific solution to the differential equation.
  • Application: By specifying an initial condition, it ensures that the solution is not just a family of curves, but a single, unique solution.
This setup is necessary because differential equations can have many solutions, and the initial condition helps to find the particular solution that fits the given criteria.
Differential Equations
Differential equations are mathematical equations that involve derivatives of an unknown function. They describe how a particular quantity changes over time or space. Our exercise presents a differential equation, \( y^{\prime}=\frac{2}{x^{4}}-y^{2} \), to be solved with an initial condition.
  • Types: There are ordinary differential equations, which involve functions of a single variable, and partial differential equations, which involve multiple variables. Our equation is an ordinary differential equation (ODE) as it involves derivatives with respect to one variable, \( x \).
  • Purpose: These equations are extensively used in fields like physics, engineering, and economics to model the behavior of systems and predict future outcomes.
Solving such equations can often provide insights into the underlying mechanics of the systems they model, making them a powerful tool in both theoretical studies and practical applications.
Numerical Methods
Numerical methods are techniques used to approximate solutions to mathematical problems that cannot be solved analytically. These methods are essential for solving differential equations, especially when no closed-form solution exists.Runge-Kutta Method:- This is a powerful numerical method especially suited for solving initial value problems in ordinary differential equations (ODEs).- It offers a balance of accuracy and efficiency, especially in the fourth-order version, which is commonly used due to its stability and precision.Steps in the Runge-Kutta Method (Fourth-Order):- Define the equation \( f(x,y) = \frac{2}{x^{4}} - y^{2} \), where the derivative and the function itself are involved.- Implement a loop that uses this method to iteratively update the \( y \) value.- Use small increments in \( x \) to precisely trace the curve of the solution. By systematically applying these steps, the method approximates the solution over a range of values, ideally capturing where zero crossings occur, as was needed in our exercise.

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Most popular questions from this chapter

Use the fourth-order Runge-Kutta subroutine with h = 0.1 to approximate the solution to $$ y^{\prime}=\cos (5 y)-x, \quad y(0)=0 $$ at the points x = 0, 0.1, 0.2, . . . , 3.0. Use your answers to make a rough sketch of the solution on 30, 34 .

An object of mass 100 kg is released from rest from a boat into the water and allowed to sink. While gravity is pull- ing the object down, a buoyancy force of 1>40 times the weight of the object is pushing the object up (weight = mg). If we assume that water resistance exerts a force on the object that is proportional to the velocity of the object, with proportionality constant 10 N-sec/m, find the equation of motion of the object. After how many sec- onds will the velocity of the object be 70 m/sec?

Early Monday morning, the temperature in the lecture hall has fallen to $$40^{\circ} \mathrm{F}$$, the same as the temperature outside. At 7:00 a.m., the janitor turns on the furnace with the thermostat set at $$70^{\circ} \mathrm{F}$$. The time constant for the building is 1>K = 2 hr and that for the building along with its heating system is 1>K1 = 1>2 hr. Assuming that the outside temperature remains constant, what will be the temperature inside the lecture hall at 8:00 a.m.? When will the temperature inside the hall reach $$65^{\circ} \mathrm{F}$$?

The solution to the initial value problem $$ \frac{d y}{d x}=(x+y+2)^{2}, \quad y(0)=-2 $$ crosses the x-axis at a point in the interval 30, 1.44 . By experimenting with the improved Euler's method subroutine, determine this point to two decimal places.

Chemical Reactions. The reaction between nitrous oxide and oxygen to form nitrogen dioxide is given by the balanced chemical equation \( 2 \mathrm{NO}+\mathrm{O}_{2}=2 \mathrm{NO}_{2} \). At high temperatures the dependence of the rate of this reaction on the concentrations of \( \mathrm{NO}, \mathrm{O}_{2} \), and \( \mathrm{NO}_{2} \) is complicated. However, at \( 25^{\circ} \mathrm{C} \) the rate at which \( \mathrm{NO}_{2} \) is formed obeys the law of mass action and is given by the rate equation $$ \frac{d x}{d t}=k(\alpha-x)^{2}\left(\beta-\frac{x}{2}\right) $$ where \( x(t) \) denotes the concentration of NO2 at time t, k is the rate constant, a is the initial concentration of NO, and \( \beta \) is the initial concentration of \( \mathrm{O}_{2} . \text { At } 25^{\circ} \mathrm{C} \), the constant k is \( 7.13 \times 10^{3} \text { (liter) }^{2} /(\text { mole })^{2}(\text { second }) \). Let \( \alpha=0.0010 \text { mole } / \mathrm{L}, \quad \beta=0.0041 \mathrm{mole} / \mathrm{L} \). Use the fourth-order Runge- Kutta algorithm to approximate \( x(10) \). For a tolerance of e = 0.000001, use a stopping procedure based on the relative error.

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