/*! 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 each differential equation... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 26

Solve each differential equation with the given initial condition. $$ \begin{array}{l} x y^{\prime}=3 y+x^{4} \\ y(1)=7 \end{array} $$

Short Answer

Expert verified
The solution is \( y(x) = x^4 + 6x^3 \).

Step by step solution

01

Set up the equation

Start with the given differential equation: \[ x y' = 3y + x^4 \]This is a first-order linear differential equation. Our goal is to solve for the function \( y(x) \).
02

Separate variables

Divide both sides of the equation by \( x \) to isolate \( y' \) on one side:\[ y' = \frac{3y}{x} + x^3 \]This equation is now in a form where we can apply standard techniques to solve it.
03

Determine the integrating factor

For a first-order linear differential equation of the form \( y' + P(x)y = Q(x) \), we find the integrating factor, \( \mu(x) \), which is given by:\[ \mu(x) = e^{\int P(x) \, dx} \]For our equation, \( P(x) = -\frac{3}{x} \), so:\[ \mu(x) = e^{\int -\frac{3}{x} \, dx} = e^{-3\ln|x|} = x^{-3} \]
04

Multiply through by the integrating factor

Multiply the entire differential equation by the integrating factor \( x^{-3} \):\[ x^{-3} y' + x^{-3} \cdot \frac{3y}{x} = x^{-3} \cdot x^3 \]Simplifying gives:\[ x^{-3}y' + \frac{3y}{x^4} = 1 \]
05

Integrate both sides

The left side of the equation is the derivative of \( (x^{-3} y) \), so we integrate both sides:\[ \int d(x^{-3} y) = \int 1 \, dx \]This gives:\[ x^{-3} y = x + C \]
06

Solve for y

Solve the equation \( x^{-3} y = x + C \) for \( y \):\[ y = x^3 (x + C) \]\[ y = x^4 + Cx^3 \]
07

Apply initial condition

Use the initial condition \( y(1) = 7 \) to find the constant \( C \).\[ 7 = 1^4 + C(1)^3 \]\[ 7 = 1 + C \]\[ C = 6 \]
08

Write the final solution

Substitute the value of \( C \) back into the solution for \( y(x) \):\[ y(x) = x^4 + 6x^3 \]

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
A first-order linear differential equation is an equation that involves the first derivative of a function and the function itself. It is generally expressed in the form:\[\frac{dy}{dx} + P(x)y = Q(x)\]Here, \( P(x) \) and \( Q(x) \) are functions of \( x \). The term \( \frac{dy}{dx} \) represents the rate of change of \( y \) with respect to \( x \). The goal is to find the function \( y(x) \) that satisfies the equation.
  • "First-order" refers to the highest derivative in the equation being the first derivative.
  • "Linear" indicates that both the function \( y \) and its derivative \( y' \) appear linearly (not in powers higher than one).
These equations are fundamental in understanding dynamic systems where one quantity changes with respect to another, like in population growth and radioactive decay.
Integrating Factor
The integrating factor is a crucial technique used to solve first-order linear differential equations of the form:\[\frac{dy}{dx} + P(x)y = Q(x)\]The idea is to multiply the entire differential equation by an integrating factor \( \mu(x) \). This converts it into an easier form that can be integrated directly. To find the integrating factor, use:\[\mu(x) = e^{\int P(x) \, dx}\]
  • The choice of \( \mu(x) \) is based on the function \( P(x) \), which affects both terms involving \( y \) and \( \frac{dy}{dx} \).
  • This step essentially transforms the left-hand side of the equation into a derivative of a product, which simplifies the integration process.
By multiplying the original equation by the integrating factor, it creates a form where you can integrate both sides easily, leading to the solution of the differential equation.
Separation of Variables
Separation of variables is another technique used to solve differential equations, although it is not directly applicable to first-order linear differential equations without adjustments. This method involves rearranging the equation so that each variable and its corresponding differential are on separate sides of the equation.
  • Typically applied to equations in the form \( M(x)dx + N(y)dy = 0 \), where separation allows integration of each part independently.
  • After separation, both sides of the equation are integrated concerning their respective variables.
This method is powerful for solving simpler differential equations and can sometimes be adapted for linear and non-linear equations alike, by considering any hyperbole between the arrangement and solvability of the components.
Initial Condition
Initial conditions are essential in finding specific solutions to differential equations. They provide a particular value that the solution must satisfy, usually expressed as \( y(x_0) = y_0 \).
  • These conditions allow us to find the constant of integration that arises from solving the differential equation.
  • Initial conditions help ensure the solution is not just a family of functions but is uniquely determined for the given problem.
In the example equation, the initial condition given is \( y(1) = 7 \). This step is crucial because even if the differential equation is solved generally, the initial condition aids in finding the precise value of any constants, leading to the exact function that meets all requirements. This eliminates ambiguity in dynamic system models by constraining the solution to the real-world scenario described by the problem.

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 medical examiner called to the scene of a murder will usually take the temperature of the body. A corpse cools at a rate proportional to the difference between its temperature and the temperature of the room. If \(y(t)\) is the temperature (in degrees Fahrenheit) of the body \(t\) hours after the murder, and if the room temperature is \(70^{\circ},\) then \(y\) satisfies $$ \begin{aligned} y^{\prime} &=-0.32(y-70) \\ y(0) &=98.6 \text { (body temperature initially } \left.98.6^{\circ}\right) \end{aligned}$$ a. Solve this differential equation and initial condition. b. Use your answer to part (a) to estimate how long ago the murder took place if the temperature of the \text { body when it was discovered was } 80^{\circ} .

Determine whether each differential equation is separable. (Do not solve it, just find whether it's separable.) $$y^{\prime}=x+y$$

For each initial value problem, use an Euler's method graphing calculator program to find the approximate solution at the stated \(x\) -value, using 50 segments. [Hint: Use an interval that begins at the initial \(x\) -value and ends at the stated \(x\) -value. \(y^{\prime}=x e^{-y}\) \(y(1)=0.5\) Approximate the solution at \(x=3\)

Solve each differential equation and initial condition and verify that your answer satisfies both the differential equation and the initial condition. $$\left\\{\begin{array}{l} y^{\prime}=y^{2} e^{x}+y^{2} \\ y(0)=1 \end{array}\right.$$

The burning of coal and oil is increasing the amount of carbon dioxide in the atmosphere, which is expected to trap more solar radiation and increase global temperatures (the "greenhouse effect"), possibly raising the sea level by melting polar ice and flooding coastal regions. The current 3300 billion tons of carbon dioxide in the atmosphere is growing by about 50 billion tons per year, while about \(1 \%\) of the accumulated carbon dioxide is removed by natural processes. Therefore, the amount \(y(t)\) of carbon dioxide (in billions of tons) \(t\) years from now satisfies $$ \begin{array}{l} y^{\prime}=-0.01 y+50 \\ y(0)=3300 \end{array} $$ a. Solve this differential equation and initial condition. b. Find when the amount of carbon dioxide will reach 4000 billion tons, at which time global temperatures are expected to have risen by 3 degrees Fahrenheit c. Use your solution to find the long-run amount of carbon dioxide in the atmosphere.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

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.