/*! 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 6 Determine whether the given diff... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Determine whether the given differential equation is exact. If it is exact, solve it. $$ \left(2 y-\frac{1}{x}+\cos 3 x\right) \frac{d y}{d x}+\frac{y}{x^{2}}-4 x^{3}+3 y \sin 3 x=0 $$

Short Answer

Expert verified
The differential equation is exact and the implicit solution is \(-\frac{y}{x} - x^4 - \frac{y}{3} \cos 3x = C\).

Step by step solution

01

Identify the Form

The given differential equation is \( M(x, y) + N(x, y) \frac{dy}{dx} = 0 \). Comparing with the standard form, we identify \( M(x, y) = \frac{y}{x^2} - 4x^3 + 3y \sin 3x \) and \( N(x, y) = 2y - \frac{1}{x} + \cos 3x \).
02

Compute Partial Derivatives

Compute \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \). We have \( \frac{\partial M}{\partial y} = 3\sin 3x \) and \( \frac{\partial N}{\partial x} = 3\sin 3x \).
03

Check Exactness Condition

The differential equation is exact if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). Since \( 3\sin 3x = 3\sin 3x \), the equation is exact.
04

Solve for the Potential Function \( \psi(x, y) \)

Integrate \( M(x, y) \) with respect to \( x \) to find \( \psi(x, y) \). From \( M(x, y) = \frac{y}{x^2} - 4x^3 + 3y \sin 3x \), integrating term-by-term gives: \( \psi(x, y) = -\frac{y}{x} - x^4 + \int 3y \sin 3x \, dx + g(y) \).
05

Find \( g(y) \) by Integrating \( N(x, y) \)

Differentiate \( \psi(x, y) \) with respect to \( y \) to find any missing \( g(y) \). \( \frac{\partial \psi}{\partial y} = -\frac{1}{x} + 3 \sin 3x + g'(y) \). Compare with \( N(x, y) \), we find \( g'(y) = 0 \) which implies \( g(y) = C \).
06

Formulate the Solution \( \psi(x, y) = C \)

Combine results: \( -\frac{y}{x} - x^4 + y (-\frac{1}{3}) \cos 3x = C \). Rearrange: \(-\frac{y}{x} - x^4 - \frac{y}{3} \cos 3x = C\). This is the implicit solution for the exact 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 91Ó°ÊÓ!

Key Concepts

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

Partial Derivatives
Understanding partial derivatives is crucial in analyzing multivariable functions, which depend on multiple inputs. In the context of exact differential equations, we use partial derivatives to check for exactness.
To determine exactness, we need to identify the functions \( M(x, y) \) and \( N(x, y) \) in the differential equation.
After that, we compute the partial derivative of \( M(x, y) \) with respect to \( y \), which gives insight into how \( M \) changes with changes in \( y \), treating \( x \) as a constant.
Next, we find the partial derivative of \( N(x, y) \) with respect to \( x \), which reflects how \( N \) varies with changes in \( x \), keeping \( y \) constant.

These derivatives serve as a check for exactness. If the partial derivatives \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \) are equal, the differential equation is exact.
This means there exists a potential function \( \psi(x, y) \) such that \( M(x, y) = \frac{\partial \psi}{\partial x} \) and \( N(x, y) = \frac{\partial \psi}{\partial y} \).
Potential Function
The potential function \( \psi(x, y) \) acts as the foundational solution for exact differential equations.
It is constructed so that its partial derivatives with respect to \( x \) and \( y \) give the original functions \( M(x, y) \) and \( N(x, y) \) of the differential equation.
To determine \( \psi(x, y) \), we integrate \( M(x, y) \) with respect to \( x \), incorporating all known terms while treating \( y \) as a constant.
This process gives a preliminary form of \( \psi \), but it might include an unknown function \( g(y) \) which accounts for terms dependent only on \( y \).
In the example, \( \int M(x, y) \, dx \) results in parts of \( \psi \), while \( g(y) \) is determined by aligning \( \frac{\partial \psi}{\partial y} \) to equal \( N(x, y) \).
The ultimate goal is to express the solution in the form \( \psi(x, y) = C \), where \( C \) is a constant. This implicitly represents the relationship between \( x \) and \( y \) that satisfies the original differential equation.
Integrating Factor
An integrating factor is a function that, when multiplied to a differential equation, makes it easier to solve by converting it into an exact equation.
Though not directly used in this exercise because the equation was already exact, it is a vital tool when the equation is not initially exact.
Finding an integrating factor might involve trial and error or more systematic approaches, such as assuming it to be a function of \( x \), \( y \), or both.
The procedure involves:
  • Multiplying the entire differential equation by the integrating factor.
  • Reassessing the partial derivatives \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \).
  • Checking if this transformation has resulted in an exact equation.
Once the equation is exact, you can find the potential function as discussed previously.
Implicit Solution
An implicit solution is a form of solution where the relationship between variables is not explicitly solved for one of the variables.
This means the solution is written as an equation involving both \( x \) and \( y \), often of the form \( \psi(x, y) = C \), where \( C \) is a constant.
In the exercise, the implicit solution is derived from the potential function \( \psi(x, y) \), and involves terms from both \( x \) and \( y \).
The resulting implicit formula can sometimes be more convenient, especially when an explicit solution would be complicated or impossible to express.
  • It provides a broader characterization of the solution space.
  • Implicit solutions are essential for representing complex relationships in multivariable calculus.
Adjusting \( C \) leads to different curves or surfaces in the solution space, depicting various possible solutions depending on initial or boundary conditions.

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 given differential equation by using an appropriate substitution. $$ 3\left(1+t^{2}\right) \frac{d y}{d t}=2 t y\left(y^{3}-1\right) $$

The differential equation $$ \frac{d y}{d x}=\frac{-x+\sqrt{x^{2}+y^{2}}}{y} $$ describes the shape of a plane curve \(C\) that will reflect all incoming light beams to the same point and could be a model for the mirror of a reflecting telescope, a satellite antenna, or a solar collector. See Problem 29 in Exercises \(1.3\). There are several ways of solving this \(\mathrm{DE}\). (a) Verify that the differential equation is homogeneous (see Section 2.5). Show that the substitution \(y=u x\) yields $$ \frac{u d u}{\sqrt{1+u^{2}}\left(1-\sqrt{\left.1+u^{2}\right)}\right.}=\frac{d x}{x}. $$ Use a CAS (or another judicious substitution) to integrate the left-hand side of the equation. Show that the curve \(C\) must be a parabola with focus at the origin and is symmetric with respect to the \(x\) -axis. (b) Show that the first differential equation can also be solved by means of the substitution \(u=x^{2}+y^{2}\).

A large tank is filled to capacity with 500 gallons of pure water. Brine containing 2 pounds of salt per gallon is pumped into the tank at a rate of \(5 \mathrm{gal} / \mathrm{min}\). The well-mixed solution is pumped out at the same rate. Find the number \(A(t)\) of pounds of salt in the tank at time \(t\).

Find the general solution of the given differential equation. Give the largest interval over which the general solution is defined. Determine whether there are any transient terms in the general solution. $$ y d x-4\left(x+y^{6}\right) d y=0 $$

The radioactive isotope of lead, \(\mathrm{Pb}-209\), decays at a rate proportional to the amount present at time \(t\) and has a half-life of \(3.3\) hours. If 1 gram of this isotope is present initially, how long will it take for \(90 \%\) of the lead to decay?

See all solutions

Recommended explanations on Physics 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.