/*! 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 6 Explain why the graph of the sol... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 6

Explain why the graph of the solution to the initial value problem \(y^{\prime}(t)=\frac{t^{2}}{1-t}, y(-1)=\ln 2\) cannot cross the line \(t=1\)

Short Answer

Expert verified
Question: Prove that the graph of the solution to the initial value problem \(y'(t) = \frac{t^2}{1-t}, y(-1)=\ln 2\) cannot cross the line \(t=1\). Answer: The graph of the given IVP's solution cannot cross the line \(t=1\) because the solution \(y(t)\) is undefined at \(t=1\), specifically due to the presence of \(\ln|1-t|\) in the solution, which is undefined when \(t=1\).

Step by step solution

01

Integration of the given equation

Integrate \(y'(t)\) to get the general solution for \(y(t)\): \(y'(t) = \frac{t^2}{1-t}\) Integrate both sides w.r.t \(t\): \(y(t) = \int \frac{t^2}{1-t} dt\)
02

Perform substitution for solving integration

Let's use substitution: \(u = 1 - t\), then \(-du = dt\): We notice that the original integral becomes: \(y(t) = -\int \frac{(1-u)^2}{u} du\)
03

Expand the integral and perform integration

Expand \((1-u)^2\) and rewrite the integral: \(y(t) = -\int \frac{(1 - 2u + u^2)}{u} du = -\int \frac{1}{u} du + 2\int du - \int u du\) Now integrate each part: \(y(t) = -\ln|u| + 2u - \frac{1}{2}u^2 + C\)
04

Re-substitute with our original variable and determine C

Replace \(u\) with the original term \(1-t\) and use the initial condition \(y(-1)=\ln 2\): \(y(t) = -\ln|1-t| + 2(1-t) - \frac{1}{2}(1-t)^2 + C\) Now plug in initial condition \(t = -1\) and \(y(-1) = \ln 2\): \(\ln 2 = -\ln|1-(-1)| + 2(1-(-1)) - \frac{1}{2}(1+1)^2 + C\) Solving for C, we get: \(C = -2\)
05

Final solution and analysis on the line \(t=1\)

Now we have the final solution: \(y(t) = -\ln|1-t| + 2(1-t) - \frac{1}{2}(1-t)^2 - 2\) Now, let's analyze the solution for the line \(t=1\): \(y(t) = -\ln|1-1| + 2(1-1) - \frac{1}{2}(1-1)^2 - 2 = -\ln|0|\) As \(\ln|0|\) is undefined, the graph of the solution cannot cross the line \(t=1\).

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.

Understanding Differential Equations
Differential equations are equations that involve an unknown function and its derivatives. They are essential tools in modeling real-world phenomena where quantities continuously change. In a differential equation like \( y'(t) = \frac{t^2}{1-t} \), \( y'(t) \) represents the derivative of \( y(t) \) with respect to \( t \), essentially denoting the rate of change of \( y \).
The main goal is to find the function \( y(t) \) that satisfies this relationship. This process often involves techniques such as separation of variables or substitution methods to simplify and integrate the equation.
Exploring Integration Techniques
Integration is a crucial technique in solving differential equations, as it helps in finding the original function from its derivative. In the given problem, we needed to integrate \( y'(t) = \frac{t^2}{1-t} \) to determine the function \( y(t) \).
  • We start by using a substitution method: let \( u = 1-t \), so \( -du = dt \).
  • The integral transforms as \( -\int \frac{(1-u)^2}{u} \, du \).
  • Expand and integrate each term separately: \( -\int \frac{1}{u} \, du + 2\int du - \int u \, du \).
This process results in the function \( y(t) \), highlighting the importance of choosing an effective integration technique and working through each step methodically.
The Role of Initial Conditions
An initial condition is a given value at a specific point that helps determine the constant of integration and find a unique solution to a differential equation. In the exercise, the initial condition is \( y(-1) = \ln 2 \).
  • This condition allows us to substitute known values into our integrated solution to solve for the constant \( C \).
  • Applying \( t = -1 \) in the integrated form helps narrow down the infinite solutions to one precise function.
Without initial conditions, we would only be able to produce a general solution, lacking the specificity required to model a particular scenario accurately.
Understanding Logarithmic Functions in Integration
Logarithmic functions often appear in integration, especially when dealing with expressions involving fractions. In this problem, the term \( -\ln|1-t| \) emerged naturally from the integration of \( \frac{1}{u} \). Here's why this matters:
  • Logarithmic expressions are crucial as they indicate undefined points, such as \( \ln|0| \), which helps explain why the solution can't cross certain lines (e.g., \( t=1 \)) because \( \ln(0) \) is undefined.
  • They serve as natural checks within the problem, ensuring solutions respect domain constraints.
Mastering how to navigate logarithmic terms in integration strengthens one's ability to solve differential equations efficiently.

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

In this section, several models are presented and the solution of the associated differential equation is given. Later in the chapter, we present methods for solving these differential equations. The delivery of a drug (such as an antibiotic) through an intravenous line may be modeled by the differential equation \(m^{\prime}(t)+k m(t)=I,\) where \(m(t)\) is the mass of the drug in the blood at time \(t \geq 0, k\) is a constant that describes the rate at which the drug is absorbed, and \(I\) is the infusion rate. a. Show by substitution that if the initial mass of drug in the blood is zero \((m(0)=0),\) then the solution of the initial value problem is \(m(t)=\frac{I}{k}\left(1-e^{-k t}\right)\) b. Graph the solution for \(I=10 \mathrm{mg} / \mathrm{hr}\) and \(k=0.05 \mathrm{hr}^{-1}\) c. Evaluate \(\lim _{t \rightarrow \infty} m(t),\) the steady-state drug level, and verify the result using the graph in part (b).

Solution of the logistic equation Use separation of variables to show that the solution of the initial value problem $$P^{\prime}(t)=r P\left(1-\frac{P}{K}\right), \quad P(0)=P_{0}$$is \(P(t)=\frac{K}{\left(\frac{K}{P_{0}}-1\right) e^{-n}+1}\)

In this section, several models are presented and the solution of the associated differential equation is given. Later in the chapter, we present methods for solving these differential equations. One possible model that describes the free fall of an object in a gravitational field subject to air resistance uses the equation \(v^{\prime}(t)=g-b v,\) where \(v(t)\) is the velocity of the object for \(t \geq 0, g=9.8 \mathrm{m} / \mathrm{s}^{2}\) is the acceleration due to gravity, and \(b>0\) is a constant that involves the mass of the object and the air resistance. a. Verify by substitution that a solution of the equation, subject to the initial condition \(v(0)=0,\) is \(v(t)=\frac{g}{b}\left(1-e^{-b t}\right)\) b. Graph the solution with \(b=0.1 \mathrm{s}^{-1}\) c. Using the graph in part (b), estimate the terminal velocity \(\lim _{t \rightarrow \infty} v(t)\)

For the following separable equations, carry out the indicated analysis. a. Find the general solution of the equation. b. Find the value of the arbitrary constant associated with each initial condition. (Each initial condition requires a different constant.) c. Use the graph of the general solution that is provided to sketch the solution curve for each initial condition. $$y^{\prime}(t)=\frac{t^{2}}{y^{2}+1} ; y(-1)=1, y(0)=0, y(-1)=-1$$

A fish hatchery has 500 fish at \(t=0,\) when harvesting begins at a rate of \(b>0\) fish/year. The fish population is modeled by the initial value problem \(y^{\prime}(t)=0.01 y-b\) \(y(0)=500,\) where \(t\) is measured in years. a. Find the fish population, for \(t \geq 0,\) in terms of the harvesting rate \(b\) b. Graph the solution in the case that \(b=40\) fish/year. Describe the solution. c. Graph the solution in the case that \(b=60\) fish/year. Describe the solution.
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Geometry

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.