/*! 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 41 (a) Find an explicit solution of... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 41

(a) Find an explicit solution of the initial-value problem $$ \frac{d y}{d x}=\frac{2 x+1}{2 y}, \quad y(-2)=-1 $$ (b) Use a graphing utility to plot the graph of the solution in part (a). Use the graph to estimate the interval \(I\) of definition of the solution. (c) Determine the exact interval \(I\) of definition by analytical methods.

Short Answer

Expert verified
(a) \( y = \pm \sqrt{x^2 + x - 1} \). (b) and (c) Interval: \( x \leq \frac{-1-\sqrt{5}}{2} \) or \( x \geq \frac{-1+\sqrt{5}}{2} \).

Step by step solution

01

Separate Variables

Start with the given differential equation \( \frac{dy}{dx} = \frac{2x+1}{2y} \). Separate the variables by multiplying both sides by \(2y\;dx\):\[ 2y \; dy = (2x+1) \; dx \]
02

Integrate Both Sides

Integrate both sides separately:\[ \int 2y \, dy = \int (2x+1) \, dx \]The integration gives:\[ y^2 = x^2 + x + C \]
03

Apply Initial Condition

Use the initial condition \( y(-2) = -1 \) to solve for \( C \). Substitute \( x = -2 \) and \( y = -1 \) into the equation:\[ (-1)^2 = (-2)^2 + (-2) + C \]\[ 1 = 4 - 2 + C \]\[ C = -1 \]
04

Express the Explicit Solution

The explicit solution with the constant \( C \) is:\[ y^2 = x^2 + x - 1 \]Solving for \( y \), we get:\[ y = \pm \sqrt{x^2 + x - 1} \]
05

Graph the Solution

Use a graphing utility to plot \( y = \pm \sqrt{x^2 + x - 1} \). Note that the graph only includes real numbers when the expression under the square root is non-negative: \( x^2 + x - 1 \geq 0 \).
06

Estimate Interval of Definition from Graph

The graph indicates that the solution is defined for values of \( x \) where \( x^2 + x - 1 \geq 0 \). This visually appears to be outside the roots of \( x^2 + x - 1 = 0 \).
07

Solve for Interval of Definition Analytically

Find the roots of the quadratic equation \( x^2 + x - 1 = 0 \) using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \quad \text{where } a=1, b=1, c=-1 \]\[ x = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \]The roots are \( x = \frac{-1 + \sqrt{5}}{2} \) and \( x = \frac{-1 - \sqrt{5}}{2} \). Thus, the interval of definition where \( x^2 + x - 1 \geq 0 \) is \( x \leq \frac{-1 - \sqrt{5}}{2} \cup x \geq \frac{-1 + \sqrt{5}}{2} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Separation of Variables
Separation of variables is a technique used to solve differential equations. It involves rearranging the equation to isolate all terms with the dependent variable on one side and all terms with the independent variable on the other. For example, consider the differential equation \( \frac{dy}{dx} = \frac{2x+1}{2y} \). By multiplying both sides by \(2y\) and \(dx\), we get the separated form:
  • \(2y \, dy = (2x+1) \, dx\)
This format allows each side to be integrated separately, simplifying the equation into a form that is more easily solvable. The integration results provide an implicit solution. Applying the initial conditions then helps find an explicit solution by determining any constant of integration.
Quadratic Equation
A quadratic equation is a second-order polynomial equation in a single variable, typically expressed in the form \(ax^2 + bx + c = 0\). Solving such equations often requires the quadratic formula:
  • \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula provides solutions for \(x\) based on the coefficients \(a\), \(b\), and \(c\). For our example, we have the equation \(x^2 + x - 1 = 0\) related to the definition interval of the explicit solution. Using the quadratic formula:
  • \[x = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}\]
These roots are crucial for determining where the expression under a radical is valid, ensuring real solutions are graphically represented.
Interval of Definition
The interval of definition is the set of all \(x\)-values for which the function is defined. For differential equations, this interval is determined by understanding where all components under operations such as square roots provide valid, real-number results. For example, given the solution \(y = \pm \sqrt{x^2 + x - 1}\), we examine the inequality:
  • \(x^2 + x - 1 \geq 0\)
To solve this, refer to the roots found from the quadratic equation. The function is defined where \(x^2 + x - 1\) is non-negative, which is:
  • \(x \leq \frac{-1 - \sqrt{5}}{2} \cup x \geq \frac{-1 + \sqrt{5}}{2}\)
Thus, these values determine the valid range of \(x\) for the solution.
Graphing Solutions
Graphing solutions of differential equations provides a visual representation of their behavior over an interval. Using tools like graphing utilities can help pinpoint intervals where solutions are valid. For the function \(y = \pm \sqrt{x^2 + x - 1}\), you would only graph points where \(x^2 + x - 1 \geq 0\). Below this region in the graph, the solution is undefined as the expression under the square root becomes negative. Looking at the graph, it becomes evident that the solution includes open intervals outside the roots, confirming:
  • \(x \leq \frac{-1 - \sqrt{5}}{2} \cup x \geq \frac{-1 + \sqrt{5}}{2}\)
Utilizing graphing utilities confirms analytical findings, sharpening understanding of how the solution behaves over different intervals.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose \(P(x)\) is continuous on some interval \(I\) and \(a\) is a number in \(I\). What can be said about the solution of the initial-value problem \(y^{\prime}+P(x) y=0, y(a)=0 ?\)

Suppose a small cannonball weighing \(16 \mathrm{lb}\) is shot vertically upward with an initial velocity \(v_{0}=300 \mathrm{ft} / \mathrm{s}\). The answer to the question, "How high does the cannonball go?" depends on whether we take air resistance into account. (a) Suppose air resistance is ignored. If the positive direction is upward, then a model for the state of the cannonball is given by \(d^{2} s / d t^{2}=-g\) (equation (12) of Section 1.3). Since \(d s / d t=v(t)\) the last differential equation is the same as \(d v / d t=-g\), where we take \(g=32 \mathrm{ft} / \mathrm{s}^{2} .\) Find the velocity \(v(t)\) of the cannonball at time \(t\). (b) Use the result obtained in part (a) to determine the height \(s(t)\) of the cannonball measured from ground level. Find the maximum height attained by the cannonball.

When interest is compounded continuously, the amount of money increases at a rate proportional to the amount \(S\) present at time \(t\), that is, \(d S / d t=r S\), where \(r\) is the annual rate of interest. (a) Find the amount of money accrued at the end of 5 years when \(\$ 5000\) is deposited in a savings account drawing \(5 \frac{3}{4} \%\) annual interest compounded continuously. (b) In how many years will the initial sum deposited bave doubled? (c) Use a calculator to compare the amount obtained in part (a) with the amount \(S=5000\left(1+\frac{1}{4}(0.0575)\right)^{5(4)}\) that is arcrued when interest is compounded quarterly.

Use a numerical solver and Euler's method to obtain a four-decimal approximation of the indicated value. First use \(h=0.1\) and then use \(h=0.05\). \(-y\) $$ y^{\prime}=x^{2}+y^{2}, \quad y(0)=1 ; y(0.5) $$

When forgetfulness is taken into account, the rate of memorization of a subject is given by $$ \frac{d A}{d t}=k_{1}(M-A)-k_{2} A $$ where \(k_{1}>0, k_{2}>0, A(t)\) is the amount to be memorized in time \(t, M\) is the total amount to be memorized, and \(M-A\) is the amount remaining to be memorized. See Problems 25 and 26 in Exercises \(1.3\). (a) Since the DE is autonomous, use the phase portrait concept of Section \(2.1\) to find the limiting value of \(A(t)\) as \(t \rightarrow \infty\). Interpret the result. (b) Solve for \(A(t)\) subject to \(A(0)=0\). Sketch the graph of \(A(t)\) and verify your prediction in part (a).
See all solutions

Recommended explanations on Physics Textbooks

Oscillations

Read Explanation

Dynamics

Read Explanation

Work Energy and Power

Read Explanation

Electrostatics

Read Explanation

Magnetism

Read Explanation

Waves Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.