/*! 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 18 Solve the given equation using a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 18

Solve the given equation using an integrating factor. Take \(t>0\). $$e^{t} y^{\prime}+y=1$$

Short Answer

Expert verified
The solution is \( y = -\frac{1}{2} e^{-t} + Ce^t \).

Step by step solution

01

Rewrite the Equation in Standard Form

Rewrite the given differential equation in the form \[ y^{\prime} + p(t) y = q(t) \] Here, divide both sides by \( e^t \) to get: \[ y^{\prime} + e^{-t} y = e^{-t} \]
02

Identify the Integrating Factor

The integrating factor \( \mu(t) \) is given by \[ \mu(t) = e^{\int p(t) \, dt} \] Here, \( p(t) = e^{-t} \). So, \[ \mu(t) = e^{\int e^{-t} \, dt} = e^{-t} \]
03

Multiply by the Integrating Factor

Multiply both sides of the differential equation by the integrating factor \( e^{-t} \): \[ e^{-t} y^{\prime} + e^{-t} e^{-t} y = e^{-t} e^{-t} \] which simplifies to: \[ e^{-t} y^{\prime} + e^{-2t} y = e^{-2t} \]
04

Simplify and Integrate

Notice that the left-hand side is the derivative of \( e^{-t} y \): \[ (e^{-t} y)^{\prime} = e^{-2t} \] Integrate both sides with respect to \( t \): \[ \int (e^{-t} y)^{\prime} \, dt = \int e^{-2t} \, dt \] which yields: \[ e^{-t} y = -\frac{1}{2} e^{-2t} + C \]
05

Solve for y

Multiply both sides of the equation by \( e^t \) to solve for \( y \): \[ y = -\frac{1}{2} e^{-t} + Ce^t \] where \( C \) is the constant of integration.

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.

Differential Equations
Differential equations are mathematical equations that involve unknown functions and their derivatives. They play a crucial role in various fields like physics, engineering, and economics. Ordinary differential equations (ODEs) contain one independent variable and its derivatives, while partial differential equations (PDEs) involve multiple independent variables. Simple examples include \(y' = y \) or \(y'' + y = 0\). Understanding differential equations helps in modeling and solving real-world problems where change is involved, such as population growth or electrical circuits.
Integrating Factor
The integrating factor is a function that simplifies the process of solving linear differential equations. It's particularly useful for equations of the form \(y' + p(t) y = q(t)\). The integrating factor is given by \( \mu(t) = e^{ \int p(t) \, dt }\). By multiplying the original differential equation by this integrating factor, the left-hand side becomes the derivative of a product, making it easier to solve through integration. For example, if we have \(y' + e^{-t} y = e^{-t}\), the integrating factor is \ ( \mu(t) = e^{-t} \ ).
Solving Linear Differential Equations
Solving linear differential equations involves a systematic approach. For an equation like \(y' + p(t) y = q(t)\), we follow these steps:
  • Rewrite the Equation: Ensure it is in the standard form.
  • Identify the Integrating Factor: Calculate \( \mu(t) = e^{\int p(t) \, dt } \).
  • Multiply Both Sides: Use the integrating factor to transform the equation.
  • Integrate: The left side becomes a derivative of a product, which can be integrated easily.
  • Solve for the Unknown Function: Perform algebraic manipulations to isolate \( y \).
For instance, rewriting the differential equation \( e^t y' + y = 1 \) and following the steps above leads us to the solution \( y = -\frac{1}{2} e^{-t} + Ce^t \).
Calculus in Differential Equations
Calculus provides the key tools for solving and understanding differential equations.
  • Integration: Used to find the antiderivative and solve the differential equations after simplifying them. For example, \( \int (e^{-t} y)' dt = \int e^{-2t} dt \) helps solve the simplified form.
  • Differentiation: Helps in understanding the behavior of functions and finding their rates of change.
  • Continuous Functions: Ensures smoother solutions and helps in making the functions interpretable.
Mastery of calculus principles is fundamental for efficiently solving differential equations and interpreting their solutions in real-world contexts.

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

Dialysis and Creatinine Clearance According to the National Kidney Foundation, in 1997 more than 260,000 Americanssuffered from chronic kidney failure and needed an artificial kidney (dialysis) to stay alive. (Source: The National Kidney Foundation, wriv.kidney.org.) When the kidneys fail, toxic waste products such as creatinine and urea build up in the blood. One way to remove these wastes is to use a process known as peritoneal dialysis, in which the paticnt's peritoneum, or lining of the abdomen, is used as a filter. When the abdominal cavity is filled with a certain dialysate solution, the waste products in the blood filter through the peritoneum into the solution. After a waiting period of several hours, the dialysate solution is drained out of the body along with the waste products. In one dialysis session, the abdomen of a patient with an elevated concentration of creatinine in the blood equal to 110 grams per liter was filled with two liters of a dialysate (containing no creatinine). Let \(f(t)\) denote the concentration of creatinine in the dialysate at time \(t .\) The rate of change of \(f(t)\) is proportional to the difference between 110 (the maximum concentration that can be attained in the dialysate) and \(f(t)\) Thus, \(f(t)\) satisfies the differential equation $$ y^{\prime}=k(110-y) $$ (a) Suppose that, at the end of a 4 -hour dialysis session, the concentration in the dialysate was 75 grams per liter and it was rising at the rate of 10 grams per liter per hour. Find \(k.\) (b) What is the rate of change of the concentration at the beginning of the dialysis session? By comparing with the rate at the end of the session, can you give a (simplistic) justification for draining and replacing the dialysate with a fresh solution after 4 hours of dialysis? [Hint: You do not need to solve the differential equation.]

Review concepts that are important in this section. In each exercise, sketch the graph of a function with the stated properties. Domain: \(0 \leq t \leq 7 ;(0,2)\) is on the graph; the slope is always positive, the slope becomes more positive as \(t\) increases from 0 to 3, and the slope becomes less positive as \(t\) increases from 3 to 7.

Radium 226 is a radioactive substance with a decay constant .00043. Suppose that radium 226 is being continuously added to an initially empty container at a constant rate of 3 milligrams per year. Let \(P(t)\) denote the number of grams of radium 226 remaining in the container after years (a) Find an initial-value problem satisfied by \(P(t).\) (b) Solve the initial-value problem for \(P(t).\) (c) What is the limit of the amount of radium 226 in the container as \(t\) tends to infinity?

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 \(f(t).\)

Rod When a red-hot steel rod is plunged in a bath of water that is kept at a constant temperature \(10^{\circ} \mathrm{C},\) the temperature of the rod at time \(t, f(t),\) satisfies the differential equation $$y^{\prime}=k[10-y]$$ where \(k>0\) is a constant of proportionality. Determine \(f(t)\) if the initial temperature of the rod is \(f(0)=350^{\circ} \mathrm{C}\) and \(k=.1.\)
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical 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.