/*! 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 26 Solve the initial-value problem.... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 26

Solve the initial-value problem. $$t y^{\prime}-y=-1, y(1)=1, t>0$$

Short Answer

Expert verified
The solution is \( y = 1 \).

Step by step solution

01

- Identify the type of differential equation

The given differential equation is: \[ t y^{\text{prime}} - y = -1 \] This is a first-order linear differential equation.
02

- Rewrite the differential equation

Rewrite the equation in the standard form: \[ y^{\text{prime}} - \frac{y}{t} = \frac{-1}{t} \]
03

- Identify the integrating factor

To solve the differential equation, find the integrating factor, \( \mu(t) \): \[ \mu(t) = e^{\int -\frac{1}{t} \, dt} = e^{-\text{ln}|t|} = \frac{1}{t} \]
04

- Multiply through by the integrating factor

Multiply both sides of the equation by the integrating factor \( \frac{1}{t} \):\[ \frac{1}{t} y^{\text{prime}} - \frac{y}{t^2} = -\frac{1}{t^2} \]
05

- Simplify the left-hand side

Recognize that the left-hand side is the derivative of \( \frac{y}{t} \): \[ \frac{d}{dt} \left( \frac{y}{t} \right) = -\frac{1}{t^2} \]
06

- Integrate both sides

Integrate both sides with respect to \( t \): \[ \int \frac{d}{dt} \left( \frac{y}{t} \right) dt = \int -\frac{1}{t^2} dt \] \[ \frac{y}{t} = \frac{1}{t} + C \], where \( C \) is the constant of integration.
07

- Solve for \( y \)

Multiply through by \( t \) to solve for \( y \): \[ y = 1 + Ct \]
08

- Apply the initial condition

Use the initial condition \( y(1) = 1 \) to find \( C \): \[ 1 = 1 + C \cdot 1 \] \[ C = 0 \]
09

- Write the final solution

Substitute \( C = 0 \) into the general solution: \[ y = 1 \]

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.

First-order linear differential equation
The given problem provides a differential equation of the form:
$$t y^{\text{prime}} - y = -1$$
This is known as a first-order linear differential equation. Let's break it down to understand why.
A first-order linear differential equation is one where the highest derivative of the function (in this case, y) is first-order, meaning it has no higher power than one. Also, it can be written in the standard linear form as:
\[ \frac{dy}{dt} + P(t) y = Q(t) \]
Here, we have:
  • Classify the given equation correctly.
  • Rewrite in the standard form: \( y^{\text{prime}} - \frac{y}{t} = \frac{-1}{t} \).
Recognizing the standard form makes it easier to apply the method of integrating factors to solve it.
This approach is crucial for these types of equations.
Integrating factor
To solve the differential equation, an integrating factor is used.
The integrating factor simplifies the equation, making it easier to solve.
Given the standard form:
  • \( y^{\text{prime}} - \frac{y}{t} = \frac{-1}{t} \)
We identify the integrating factor, \( \mu(t) = e^{\int P(t) dt} \). Here, \( P(t) = -\frac{1}{t} \).
Calculate the integrating factor as:
  • \( \mu(t) = e^{\int -\frac{1}{t} dt} = e^{-ln|t|} = \frac{1}{t} \)
Multiply the differential equation by the integrating factor:
  • \( \frac{1}{t} y^{\text{prime}} - \frac{y}{t^2} = -\frac{1}{t^2} \)
The left-hand side now becomes the derivative of \( \frac{y}{t} \):
\( \frac{d}{dt} \left( \frac{y}{t} \right) = -\frac{1}{t^2} \)
This makes it much easier to integrate both sides.
Constant of integration
Upon integrating both sides of the simplified differential equation, we introduce the constant of integration, \( C \).
Integrating both sides with respect to \( t \):
  • \( \int \frac{d}{dt} \left( \frac{y}{t} \right) dt = \int -\frac{1}{t^2} dt \)
This results in:
  • \( \frac{y}{t} = \frac{1}{t} + C \)
The constant of integration, \( C \), represents an arbitrary constant which we need to solve for using the initial condition provided:
  • \( y(1) = 1 \)
Solving with the initial condition, we get:
  • \( 1 = 1 + C \cdot 1 \)
  • \( C = 0 \)
Finally, the solution to the differential equation is:
  • \( y = 1 \)
Understanding the constant of integration is key to solving initial-value problems accurately.

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

A person deposits \(\$ 10,000\) in a bank account and decides to make additional deposits at the rate of \(A\) dollars per year. The bank compounds interest continuously at the annual rate of \(6 \%,\) and the deposits are made continuously into the account. (a) Set up a differential equation that is satisfied by the amount \(f(t)\) in the account at time \(t.\) (b) Determine \(f(t)\) (as a function of \(A\) ). (c) Determine \(A\) if the initial deposit is to double in 5 years.

Solve the following differential equations with the given initial conditions. $$3 y^{2} y^{\prime}=-\sin t, y\left(\frac{\pi}{2}\right)=1$$

Solve the initial-value problem. $$y^{\prime}=2(10-y), y(0)=1$$

Suppose that \(f(t)\) satisfies the initial-value problem \(y^{\prime}=y^{2}+t y-7, y(0)=3 .\) Is \(f(t)\) increasing or decreasing at \(t=0 ?\)

A single dose of iodine is injected intravenously into a patient. The iodine mixes thoroughly in the blood before any is lost as a result of metabolic processes (ignore the time required for this mixing process). Iodine will leave the blood and enter the thyroid gland at a rate proportional to the amount of iodine in the blood. Also, iodine will leave the blood and pass into the urine at a (different) rate proportional to the amount of iodine in the blood. Suppose that the iodine enters the thyroid at the rate of \(4 \%\) per hour, and the iodine enters the urine at the rate of \(10 \%\) per hour. Let \(f(t)\) denote the amount of iodine in the blood at time \(t\) Write a differential equation satisfied by \(f(t).\)
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

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.