/*! 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 35 (a) Use partial fractions to sho... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 35

(a) Use partial fractions to show that $$\int \frac{d u}{u^{2}-a^{2}}=\frac{1}{2 a} \ln \left|\frac{u-a}{u+a}\right|+C$$ (b) Use your result in (a) to find a solution of $$\frac{d y}{d x}=y^{2}-4$$ that passes through (i) \((0,0)\), (ii) \((0,2)\), and (iii) \((0,4)\).

Short Answer

Expert verified
(a) The integral evaluates to \( \frac{1}{2a} \ln \left| \frac{u-a}{u+a} \right| + C \). (b) Solutions: (i) \(y=0\), (ii) No real solution, and (iii) \(y=4\).

Step by step solution

01

Setup for Partial Fraction Decomposition

The integral to evaluate is \( \int \frac{du}{u^2 - a^2} \). Recognize \( u^2 - a^2 \) is a difference of squares and factor it as \((u-a)(u+a)\). Decompose the fraction \( \frac{1}{u^2 - a^2} \) into partial fractions: \( \frac{1}{u^2 - a^2} = \frac{A}{u-a} + \frac{B}{u+a} \).
02

Determine Coefficients for Partial Fractions

Clear the denominators by multiplying through by \((u-a)(u+a)\) giving \(1 = A(u+a) + B(u-a) \). Solve for \(A\) and \(B\) by substituting specific values for \(u\). Set \(u = a\) to find \(B\), and set \(u = -a\) to find \(A\).
03

Solve for A and B

Substitute \(u = a\) to get \(1 = B(2a)\) which gives \(B = \frac{1}{2a}\). Substitute \(u = -a\) to get \(1 = A(-2a)\) and solve for \(A\), giving \(A = -\frac{1}{2a}\).
04

Write the Decomposed Form

Substitute \(A\) and \(B\) back into the partial fraction decomposition: \( \frac{1}{u^2 - a^2} = \frac{-1/2a}{u-a} + \frac{1/2a}{u+a} \).
05

Integrate the Decomposed Expression

Integrate term by term: \( \int \frac{-1/2a}{u-a} \, du + \int \frac{1/2a}{u+a} \, du \). These are standard logarithmic integrals: \( \frac{-1}{2a} \ln|u-a| + \frac{1}{2a} \ln|u+a| + C \).
06

Simplify the Result

Combine the logarithms using properties of logarithms: \[ \frac{1}{2a} \ln \left| \frac{u+a}{u-a} \right| + C = \frac{1}{2a} \ln \left| \frac{u-a}{u+a} \right| + C \]. This result matches the provided form.
07

Solve the Differential Equation Using the Result

The differential equation is \( \frac{dy}{dx} = y^2 - 4 \). Recognize this is of the form \( \frac{1}{y^2 - 4} dy = dx \). Integrate both sides: \( \int \frac{1}{y^2 - 4} \, dy = \int \, dx \). Use part (a) where \( a = 2 \), giving \( \frac{1}{4} \ln \left| \frac{y-2}{y+2} \right| = x + C \).
08

Determine Constant C for Initial Conditions

Use initial condition (i) \((0,0)\): \( \frac{1}{4} \ln \left| \frac{0-2}{0+2} \right| = 0 + C \) gives \(C = 0\) since \( \ln 1 = 0\). Initial condition (ii) \((0,2)\) leads to an undefined log equation (no real solution), and initial condition (iii) \((0,4)\): \( \frac{1}{4} \ln \left| \frac{4-2}{4+2} \right| = 0 + C \) also gives \(C = 0\).

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.

Partial Fractions
Partial fractions are a crucial method in calculus for simplifying complex rational expressions into simpler forms that are easier to integrate or differentiate. This decomposition involves expressing a rational function, such as \( \frac{1}{u^2 - a^2} \), into a sum of fractions with simpler denominators.
Let's take a closer look at the process:
  • Identify the form of the denominator: In this case, \( u^2 - a^2 \) is a difference of squares, which can be factored into \((u-a)(u+a)\).
  • Set up the partial fraction decomposition: Our goal is to express \( \frac{1}{u^2 - a^2} \) as \( \frac{A}{u-a} + \frac{B}{u+a} \).
  • Determine the coefficients: To find \( A \) and \( B \), clear the denominators by multiplying through by \( (u-a)(u+a) \), resulting in an equation that you can solve by choosing strategic values for \( u \). Setting \( u = a \) and \( u = -a \) simplifies this process, allowing us to solve for \( B \) and \( A \) respectively.
Understanding partial fractions is crucial for simplifying complicated integrals, and is foundational for solving differential equations efficiently.
Integral Calculus
Integral calculus focuses on the concept of integration, which is essentially the opposite operation to differentiation. It involves finding the total accumulation of quantities, such as area under a curve. In this exercise, we build on the partial fraction decomposition result to integrate the function.
After decomposition into partial fractions (\( \frac{-1/2a}{u-a} + \frac{1/2a}{u+a} \)), each term can be integrated separately. This is where the powerful technique of recognizing the 'standard forms' of integrals comes into play.
  • Integrating basic logarithmic forms: The terms like \( \frac{1}{u} \) integrate directly to natural logarithms. Thus, \( \int \frac{-1/2a}{u-a} \, du \) becomes \( \frac{-1}{2a} \ln|u-a| \) and similarly for the second term.
  • Applying properties of logarithms: Once you have the logarithmic sums, use logarithmic identities to combine them. For example, \( \ln|u-a| - \ln|u+a| = \ln \left| \frac{u-a}{u+a} \right| \). This simplifies the final expression significantly.
Mastering the basic forms of integrals and properties of logarithms greatly aids in solving more complicated calculus problems.
Differential Equations
Differential equations involve equations with unknown functions and their derivatives. They are prevalent in many fields such as physics, engineering, and economics. In the exercise, we tackled the differential equation \( \frac{dy}{dx} = y^2 - 4 \).
This is a separable differential equation, which means it can be expressed as \( \frac{1}{y^2 - 4} dy = dx \), allowing us to integrate both sides separately:
  • Integration using results from partial fractions: We use the previously derived integral of \( \int \frac{1}{y^2 - 4} \, dy = \frac{1}{4} \ln \left| \frac{y-2}{y+2} \right| + C \). This highlights how interconnected integral calculus and differential equations can be.
  • Finding particular solutions: Substituting initial conditions such as \( (0,0) \) and \( (0,4) \) into the integrated expression helps find specific solutions or values for \( C \). However, not all conditions provide valid real solutions, as seen with the point \( (0,2) \).
Differential equations, especially separable ones, often rely heavily on integration techniques to resolve them into meaningful solutions that correspond to real-life scenarios.

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

Use the partial-fraction method to solve $$\frac{d y}{d x}=2 y(3-y)$$ where \(y_{0}=5\) for \(x_{0}=1\).

Suppose that you follow the size of a population over time. When you plot the size of the population versus time on a semilog plot (i.e., the horizontal axis, representing time, is on a linear scale, whereas the vertical axis, representing the size of the population, is on a logarithmic scale), you find that your data fit a straight line which intercepts the vertical axis at 1 (on the log scale) and has slope \(-0.43 .\) Find a differential equation that relates the growth rate of the population at time \(t\) to the size of the population at time \(t\).

Denote the size of a population at time \(t\) by \(N(t)\), and assume that $$\frac{d N}{d t}=2 N(N-10)\left(1-\frac{N}{100}\right) \quad \text { for } t \geq 0$$ (a) Find all equilibria of \((8.72)\). (b) Use the eigenvalue approach to determine the stability of the equilibria you found in (a). (c) Set $$g(N)=2 N(N-10)\left(1-\frac{N}{100}\right)$$ for \(N \geq 0\), and graph \(g(N) .\) Identify the equilibria of \((8.72)\) on your graph, and use the graph to determine the stability of the equilibria. Compare your results with your findings in (b). Use your graph to give a graphical interpretation of the eigenvalues associated with the equilibria.

Solve each differential equation with the given initial condition. \(\frac{d y}{d x}=(y+1) e^{-x}\), with \(y_{0}=2\) if \(x_{0}=0\)

(Adapted from Crawley, 1997) Denote plant biomass by \(V\), and herbivore number by \(N .\) The plant-herbivore interaction is modeled as $$ \begin{array}{l} \frac{d V}{d t}=a V\left(1-\frac{V}{K}\right)-b V N \\ \frac{d N}{d t}=c V N-d N \end{array} $$ (a) Suppose the herbivore number is equal to \(0 .\) What differential equation describes the dynamics of the plant biomass? Can you explain the resulting equation? Determine the plant biomass equilibrium in the absence of herbivores. (b) Now assume that herbivores are present. Describe the effect of herbivores on plant biomass; that is, explain the term \(-b V N\) in the first equation. Describe the dynamics of the herbivoresthat is, how their population size increases and what contributes to decreases in their population size. (c) Determine the equilibria (1) by solving $$ \frac{d V}{d t}=0 \quad \text { and } \quad \frac{d N}{d t}=0 $$ and (2) graphically. Explain why this model implies that "plant abundance is determined solely by attributes of the herbivore," as stated in Crawley (1997).
See all solutions

Recommended explanations on Biology Textbooks

Ecosystems

Read Explanation

Cellular Energetics

Read Explanation

Astrobiological Science

Read Explanation

Control of Gene Expression

Read Explanation

Microbiology

Read Explanation

Biological Structures

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.