/*! 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 138 Solve the differential equation:... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 138

Solve the differential equation: \(\mathrm{y}^{\prime}+\mathrm{y}=\mathrm{x}\).

Short Answer

Expert verified
The general solution to the given first-order linear differential equation \(\mathrm{y}^{\prime}+\mathrm{y}=\mathrm{x}\) is given by: \(y = x - 1 + \frac{C}{e^x}\)

Step by step solution

01

Find the Integrating Factor

An integrating factor for a first-order linear differential equation is given by the exponential of the integral of the coefficient of y, which in this case is simply 1. So, we have: Integrating Factor (IF) = \(e^{\int 1 dx} = e^x\)
02

Multiply the Equation by the Integrating Factor

Now, we multiply both sides of the given equation by the integrating factor, which results in: \[e^x (y' + y) = x e^x\] Now, observe that the left side of the equation is the derivative of the product of y and the integrating factor (IF): \[\frac{d}{dx}(ye^x) = x e^x\]
03

Integrate Both Sides

We'll integrate both sides of the equation to solve for \(ye^x\): \[\int \frac{d}{dx}(ye^x) dx = \int x e^x dx\] Using integration by parts on the right side of the equation, let u = x and dv = \(e^x\) dx. Then, du = dx, and v = \(e^x\) Using the integration by parts formula, \(\int udv = uv - \int vdu\), we get: \(ye^x = xe^x - e^x + C\) Where C is the constant of integration.
04

Solve for y

Now we'll solve for y by dividing both sides of the equation by \(e^x\): \(y = x - 1 + \frac{C}{e^x}\) This is the general solution to the given differential equation.

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.

Integrating Factor
When solving first-order linear differential equations, the integrating factor is a powerful tool. It helps in simplifying the equation to make it solvable. For a differential equation in the form \(y' + py = q\), the integrating factor \(\mu(x)\) is derived using:
  • \(\mu(x) = e^{\int p\, dx}\)
In our example, the coefficient \(p\) is 1, thus:
  • \(\mu(x) = e^{\int 1\, dx} = e^x\)
This factor, \(e^x\), is multiplied throughout the equation to transform it into a simpler form. Once applied, the left-hand side becomes the derivative of a product, making it easier to handle during integration.
Integration by Parts
Integration by parts is a technique used to solve integrals by reducing them to simpler parts. It is particularly useful when dealing with the product of functions, such as \(x \times e^x\).
The formula for integration by parts is given by:
  • \(\int u \, dv = uv - \int v \, du\)
In the context of our exercise, when integrating \(xe^x\), we let:
  • \(u = x\)
  • \(dv = e^x \, dx\)
  • \(du = dx\)
  • \(v = e^x\)
Applying the formula, we find that:
  • \(\int x e^x \, dx = xe^x - \int e^x \, dx\)
  • = \(xe^x - e^x + C\)
Here, \(C\) is the constant of integration, which we discuss next.
Constant of Integration
The constant of integration, often denoted as \(C\), is a fundamental element in indefinite integrals. It accounts for the infinite number of possible functions that differ by a constant.
During integration, especially when finding general solutions to differential equations, \(C\) represents all those solutions that vary only by constant values. In our solution:
  • The expression \(ye^x = xe^x - e^x + C\) includes a \(C\) because the integration process inherently leaves room for any constant addition.
Once we solve for \(y\) by dividing through by \(e^x\), the constant \(C\) remains:
  • \(y = x - 1 + \frac{C}{e^x}\)
This incorporation of \(C\) ensures the solution encompasses all possible solutions to the differential equation related by differing constants.

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

Solve the differential equation: \(\quad \mathrm{y}^{\prime}+\mathrm{y}=\sin \mathrm{x}\)

Solve \(y^{\prime}-(3 / x) y=x^{4} y^{(1 / 3)}\)

Show that the Bernoulli's equation, \((\mathrm{dy} / \mathrm{d} \mathrm{x})+\mathrm{py}=\mathrm{Qy}^{\mathrm{n}}\), where \(\mathrm{p}\) and \(\mathrm{Q}\) are functions of \(\mathrm{x}\) only, and \(\mathrm{n}\) is a constant other than zero or unity, may be transformed into a linear equation by the substitution \(\mathrm{t}=\mathrm{y}^{1-\mathrm{n}} .\) Hence solve the equation $$ (d y / d x)=\left[\left(x^{2} y^{2}+2 y\right) / x\right] $$ If \(\mathrm{n}\) is zero or unity, how may the equation be solved?

Solve \(y\left(6 y^{2}-x-1\right) d x+2 x d y=0\)

Solve \(\left[\left(\mathrm{x}^{2} / \mathrm{y}\right)+\mathrm{y}^{2}\right] \mathrm{d} \mathrm{x}-\left[\left(\mathrm{x}^{3} / \mathrm{y}^{2}\right)+\mathrm{xy}+\mathrm{y}^{2}\right] \mathrm{dy}=0\)
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Logic and Functions

Read Explanation

Calculus

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.