/*! 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 8 Solve the first-order linear dif... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 8

Solve the first-order linear differential equation. $$ y^{\prime}+2 x y=4 x $$

Short Answer

Expert verified
The solution of the given first-order linear differential equation is \(y = 2 + C e^{-x^2}\).

Step by step solution

01

Identification of Integrating Factor

The integrating factor for a first-order linear differential equation of the form \(y'+p(x)y = q(x)\) is given by \(e^{\int{p(x)dx}}\). In this equation, \(p(x)=2x\). Therefore, the integrating factor is \(e^{\int 2x dx} = e^{x^2}.\)
02

Apply the Integrating Factor

Multiplying every term in the equation by the integrating factor \(e^{x^2}\), we get \((y e^{x^2})' = 4x e^{x^2}\). This comes from the fact that the left side of the equation becomes the derivative of \(y\) times the integrating factor.
03

Integrate Both Sides

Integrating both sides with respect to \(x\), we get \(\int(y' e^{x^2}) dx = \int(4x e^{x^2}) dx\). Solving both sides, the left side simplifies to \(y e^{x^2}\). The integral on the right side can be solved by substitution (\(u=x^2\)), yielding \(2e^{x^2} + C\), with \(C\) being the constant of integration.
04

Solve for y

Finally, solving for \(y\) gives the solution of the differential equation: \(y = 2 + C e^{-x^2}\).

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
To solve a first-order linear differential equation, we often use an integrating factor. This transforms the equation into a form that is easier to solve. A first-order linear differential equation looks like:
  • \( y' + p(x)y = q(x) \)
The integrating factor is computed using the formula:
  • \( e^{\int{p(x)dx}} \)
This factor is derived from the coefficient of \( y \), known as \( p(x) \). By multiplying every term of the differential equation by this integrating factor, the left-hand side of the equation transforms into the derivative of \( y \) multiplied by the integrating factor.
This allows us to easily integrate the transformed equation.
Integration by Substitution
To solve integrals that arise in differential equations, sometimes substitution is a straightforward method. This technique is particularly useful when the integration is not immediately evident. Consider the integral
  • \( \int 4x e^{x^2} \,dx \)
In this case, by setting \( u = x^2 \), the differential becomes \( du = 2x \, dx \), allowing for a substitution that simplifies the process. This method effectively reconfigures challenging integrals into simpler ones. The goal is to reach a more manageable form to integrate readily.
Remember to substitute back the original variable after integrating.
Constant of Integration
When integrating, it's crucial to include the constant of integration, denoted as \( C \). This constant accounts for the family of solutions that differ only by a constant value.
  • For instance, solving \( \int 4x e^{x^2} \,dx \) yields \( 2e^{x^2} + C \).
The constant of integration is essential in representing the general solution to a differential equation. In the context of solving the entire equation, like \( y = 2 + C e^{-x^2} \), \( C \) helps outline all potential solutions that satisfy the differential equation.
This highlights the importance of including \( C \) in your final answer, providing the most comprehensive 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 value of a tract of timber is \(V(t)=100,000 e^{0.8 \sqrt{t}}\) where \(t\) is the time in years, with \(t=0\) corresponding to 1998 . If money earns interest continuously at \(10 \%\), the present value of the timber at any time \(t\) is \(A(t)=V(t) e^{-0.10 t}\). Find the year in which the timber should be harvested to maximize the present value function.

It is known that \(y=C e^{k x}\) is a solution of the differential equation \(y^{\prime}=0.07 y .\) Is it possible to determine \(C\) or \(k\) from the information given? If so, find its value.

Find the orthogonal trajectories of the family. Use a graphing utility to graph several members of each family. \(x^{2}+y^{2}=C\)

Find the particular solution that satisfies the initial condition. \(\left(x \sec \frac{y}{x}+y\right) d x-x d y=0 \quad y(1)=0\)

Glucose is added intravenously to the bloodstream at the rate of \(q\) units per minute, and the body removes glucose from the bloodstream at a rate proportional to the amount present. Assume that \(Q(t)\) is the amount of glucose in the bloodstream at time \(t\). (a) Determine the differential equation describing the rate of change of glucose in the bloodstream with respect to time. (b) Solve the differential equation from part (a), letting \(Q=Q_{0}\) when \(t=0\). (c) Find the limit of \(Q(t)\) as \(t \rightarrow \infty\).
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Mechanics 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.