/*! 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 1 Find the general solution of the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the general solution of the first-order, linear equation. $$ y^{\prime}+y=2 $$

Short Answer

Expert verified
The general solution is \( y = 2 + Ce^{-x} \).

Step by step solution

01

Identify the Type of Differential Equation

The given equation, \( y' + y = 2 \), is a first-order linear differential equation. It is in the standard form \( y' + P(x)y = Q(x) \), where \( P(x) = 1 \) and \( Q(x) = 2 \).
02

Calculate the Integrating Factor

The integrating factor \( \mu(x) \) is found using the formula: \( \mu(x) = e^{\int P(x) \, dx} \). Here, \( P(x) = 1 \), so: \[ \mu(x) = e^{\int 1 \, dx} = e^x \].
03

Multiply through by the Integrating Factor

Multiply every term in the differential equation by the integrating factor \( \mu(x) = e^x \): \[ e^x y' + e^x y = 2e^x \].
04

Recognize Left-Hand Side as a Derivative

The left-hand side \( e^x y' + e^x y \) can be recognized as the derivative of \( y e^x \). Thus, \[ \frac{d}{dx}(y e^x) = 2e^x \].
05

Integrate Both Sides

Integrate both sides with respect to \( x \): \[ \int \frac{d}{dx}(y e^x) \, dx = \int 2e^x \, dx \],which gives \[ y e^x = 2e^x + C \], where \( C \) is the constant of integration.
06

Solve for \( y \)

Solve the equation \( y e^x = 2e^x + C \) for \( y \): \[ y = \frac{2e^x + C}{e^x} \],which simplifies to \[ y = 2 + Ce^{-x} \].

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 91Ó°ÊÓ!

Key Concepts

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

Understanding the Integrating Factor
When dealing with first-order linear differential equations, the integrating factor simplifies the process of finding a solution. For an equation in the form \( y' + P(x)y = Q(x) \), the integrating factor \( \mu(x) \) is given by the formula:
  • \( \mu(x) = e^{\int P(x) \, dx} \)
In the given exercise, \( P(x) = 1 \). Thus, the integrating factor is:
  • \( \mu(x) = e^{x} \)
By multiplying every term of the differential equation by this factor, you transform the equation into a form where the left-hand side becomes a derivative of a product. This enables us to recognize it as:
  • \( \frac{d}{dx}(ye^x) = 2e^x \)
This product rule recognition simplifies the equation, making it easier to solve for \( y \). It’s an efficient method to solve linear differential equations.
General Solution of the Differential Equation
The general solution provides a formula that describes all possible solutions of a differential equation. In our exercise, after manipulating the equation using the integrating factor, we reach the form:
  • \( \frac{d}{dx}(y e^x) = 2e^x \)
To find the general solution, we integrate both sides. Integrating the right side:
  • \( \int 2e^x \, dx \)
  • leads to \( 2e^x + C \)
where \( C \) is the constant of integration. The solution to the original equation is then:
  • \( y e^x = 2e^x + C \)
By dividing through by \( e^x \), we isolate \( y \):
  • \( y = 2 + Ce^{-x} \)
This general solution encapsulates all solutions due to the inclusion of the constant \( C \).
Role of the Constant of Integration
The constant of integration \( C \) appears when we integrate a function, indicating there are infinitely many possible solutions. This is because integration is essentially the reverse of differentiation, and different functions can have the same derivative. In the context of our differential equation, it represents a family of solutions.
After integrating the equation \( \int 2e^x \, dx \), we add \( C \) to express this infinite range:
  • \( y e^x = 2e^x + C \)
When solving for \( y \), the constant \( C \) remains part of the solution:
  • \( y = 2 + Ce^{-x} \)
Different values of \( C \) will determine different particular solutions within the family. This constant allows us to include any initial condition to find a specific solution.

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

The presence of nonlinear terms prevents us from using the technique of this section. In special cases, a change of variable will transform the nonlinear equation into one that is linear. The equation known as Bernoulli's equation, $$ x^{\prime}=a(t) x+f(t) x^{n}, \quad n \neq 0,1, $$ was proposed for solution by James Bernoulli in December 1695. In 1696, Leibniz pointed out that the equation can be reduced to a linear equation by taking \(x^{1-n}\) as the dependent variable. Show that the change of variable, \(z=x^{1-n}\), will transform the nonlinear Bernoulli equation into the linear equation $$ z^{\prime}=(1-n) a(t) z+(1-n) f(t) $$

A tank contains 100 gal of pure water. A salt solution with concentration \(3 \mathrm{lb} / \mathrm{gal}\) enters the tank at a rate of \(2 \mathrm{gal} / \mathrm{min}\). Solution drains from the tank at a rate of \(2 \mathrm{gal} / \mathrm{min}\). Use qualitative analysis to find the eventual concentration of the salt solution in the tank.

Sensitivity to initial conditions is well illustrated by a little target practice with your numerical solver. In Exercises \(1-12\), you are given a differential equation \(x^{\prime}=f(t, x)\) and a "target." In cach case, enter the equation into your numerical solver, and then experiment with initial conditions at the given value of \(t_{0}\) until the solution of \(x^{\prime}=f(t, x)\), with \(x\left(t_{0}, x_{0}\right)\), "hits" the given target. We will use the simple linear equation, \(x^{\prime}=x-t\). The initial conditions are at \(t_{0}=0\). The target is $$ (6,0) $$

A skydiver jumps from a plane and opens her chute. One possible model of her velocity \(v\) is given by $$ m \frac{d v}{d t}=m g-k v, $$ where \(m\) is the combined mass of the skydiver and her parachute, \(g\) is the acceleration due to gravity, and \(k\) is a proportionality constant. Assuming that \(m, g\), and \(k\) are all positive constants, use qualitative analysis to determine the skydiver's "terminal velocity."

Find the general solution of each homogeneous equation. $$ \left(x^{2}+y^{2}\right) d x-2 x y d y=0 $$

See all solutions

Recommended explanations on Math Textbooks

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.