/*! 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 9 In Problems 1-22, solve the give... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 9

In Problems 1-22, solve the given differential equation by separation of variables. $$ y \ln x \frac{d x}{d y}=\left(\frac{y+1}{x}\right)^{2} $$

Short Answer

Expert verified
\(x \ln x - x = \frac{y^2}{2} + 2y + \ln |y| + C\) is the implicit solution.

Step by step solution

01

Rearrange the Equation

The given equation is \(y \ln x \frac{dx}{dy} = \left(\frac{y+1}{x}\right)^{2}\). Start by separating the variables \(x\) and \(y\). Multiply both sides by \(\frac{dy}{y}\) to isolate the differential \(dx\). This results in \(\ln x \cdot dx = \frac{(y+1)^2}{y} \cdot\frac{dy}{x^2}\).
02

Separate Variables

Separate the variables by integrating each side with respect to its own variable. Rewrite the equation as \(x^2 \ln x \, dx = \frac{(y+1)^2}{y} \, dy\). Move all expressions in terms of \(x\) to one side and those in terms of \(y\) to the other side resulting in \(\ln x \, dx = \frac{(y+1)^2}{y} \, dy\).
03

Integrate Both Sides

Integrate both sides separately. Integrate \(\ln x \, dx\) and \(\frac{(y+1)^2}{y} \, dy\):\[ \int \ln x \, dx = \int \frac{(y+1)^2}{y} \, dy \]The integration results in separate expressions for each side.
04

Solve the Integrals

Perform integration:- The integral \(\int \ln x \, dx\) involves integration by parts and equals to \(x \ln x - x + C_1\).- For \(\int \frac{(y+1)^2}{y} \, dy\), simplify the expression to \((y + 2 + \frac{1}{y})\) which results in \(\frac{y^2}{2} + 2y + \ln |y| + C_2\).
05

Combine the Solutions

The left side integration yields \(x \ln x - x + C_1\) and the right side gives \(\frac{y^2}{2} + 2y + \ln |y| + C_2\). Combine these to form the equation:\[ x \ln x - x = \frac{y^2}{2} + 2y + \ln |y| + C \] where \(C = C_2 - C_1\) is the constant of integration.
06

Present Final Solution

The final form of the solution after separating variables and integrating is:\[ x \ln x - x = \frac{y^2}{2} + 2y + \ln |y| + C \]This implicit solution relates \(x\) and \(y\).

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.

Separation of Variables
Separation of variables is a powerful technique used to solve differential equations by dividing and isolating terms involving different variables on opposite sides of the equation. This method simplifies complex equations by allowing you to integrate each variable separately. In practice, you aim to rearrange the equation so that all terms involving one variable, say \(x\), are multiplied by the \(dx\) differential, and terms involving the other variable, say \(y\), are multiplied by the \(dy\) differential.
  • First, identify the variables and differentials within the equation. In our example, we have the differential equation: \(y \ln x \, \frac{dx}{dy} = \left(\frac{y+1}{x}\right)^2\).
  • Next, manipulate the equation to isolate these variables on either side. Multiplying both sides by \(\frac{dy}{y}\), we get: \(\ln x \cdot dx = \frac{(y+1)^2}{y} \, dy\).
  • This produces two integrals, \(\int \ln x \, dx\) and \(\int \frac{(y+1)^2}{y} \, dy\), allowing us to solve them independently.
This technique lays the groundwork for finding an implicit solution, which connects the variables in a relationship involving a constant of integration.
Integration by Parts
Integration by parts is an essential tool in calculus for solving integrals where the standard method of integration is not easily applicable. It's particularly useful when dealing with the product of functions, such as \(\int u \, dv\), by using the formula:\[\int u \, dv = uv - \int v \, du\]This formula helps reduce complex integrals into simpler ones. Here鈥檚 how it works in our problem:
  • Choose the function \(u\) (which in our case is \(\ln x\)) and \(dv\) (which is \(dx\)). Differentiate \(u\) to get \(du\) and integrate \(dv\) to get \(v\).
  • Our example integral \(\int \ln x \, dx\) is solved by selecting \(u = \ln x\) and \(dv = dx\). Then, \(du = \frac{1}{x} \, dx\) and \(v = x\).
  • Substituting these into the integration by parts formula yields \(x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - x + C_1\).
Using integration by parts transforms a challenging integral into a more manageable one, allowing us to progress further in solving the differential equation.
Implicit Solutions
Implicit solutions are forms in which the solution to a differential equation is given as a relationship between the variables rather than explicitly solving for one variable in terms of the other. This approach often emerges in complicated equations where finding an explicit solution is either difficult or impossible.In our context, after integrating both sides, we arrive at an implicit solution:\[ x \ln x - x = \frac{y^2}{2} + 2y + \ln |y| + C \]Here鈥檚 why implicit solutions are valuable:
  • They offer flexibility in solving complex relationships, especially when the equation involves nonlinear terms.
  • Since both sides of the equation have been integrated separately, you need only ensure continuity across the solution by applying initial conditions or extra constraints when available, to determine the constant \(C\).
  • Implicit solutions help describe the relationship between \(x\) and \(y\) comprehensively, even without giving a direct formula \(x = f(y)\) or \(y = g(x)\).
In problem-solving contexts, implicit solutions are both revealing and comprehensive, giving a complete picture of the relationship between the variables from the differential equation.

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 tank in the form of a right circular cylinder standing on end is leaking water through a circular hole in its bottom. As we saw in (10) of Section 1.3, when friction and contraction of water at the hole are ignored, the height \(h\) of water in the tank is described by $$ \frac{d h}{d t}=-\frac{A_{h}}{A_{w}} \sqrt{2 g h}, $$ where \(A_{w}\) and \(A_{h}\) are the cross-sectional areas of the water and the hole, respectively. (a) Solve for \(h(t)\) if the initial height of the water is \(H\). By hand, sketch the graph of \(h(t)\) and give its interval \(I\) of definition in terms of the symbols \(A_{w}, A_{h}\), and \(H\). Use \(g=32 \mathrm{ft} / \mathrm{s}^{2}\) (b) Suppose the tank is \(10 \mathrm{ft}\) high and has radius \(2 \mathrm{ft}\) and the circular hole has radius \(\frac{1}{2}\) in. If the tank is initially full, how long will it take to empty?

Solve the given initial-value problem. Give the largest interval \(I\) over which the solution is defined. $$ \left(\frac{e^{-2 \sqrt{x}}-y}{\sqrt{x}}\right) \frac{d x}{d y}=1, \quad y(1)=1 $$

A small metal bar is removed from an oven whose temperature is a constant \(300^{\circ} \mathrm{F}\) into a room whose temperature is a constant \(70^{\circ} \mathrm{F}\). Simultaneously, an identical metal bar is removed from the room and placed into the oven. Assume that time \(t\) is measured in minutes. Discuss: Why is there a future value of time, call it \(t^{*}>0\), at which the temperature of each bar is the same?

A differential equation governing the velocity \(v\) of a falling mass \(m\) subjected to air resistance proportional to the square of the instantaneous velocity is $$ m \frac{d v}{d t}=m g-k v^{2}, $$ where \(k>0\) is the drag coefficient. The positive direction is downward. (a) Solve this equation subject to the initial condition \(v(0)=v_{0}\) (b) Use the solution in part (a) to determine the limiting, or terminal, velocity of the mass. We saw how to determine the terminal velocity without solving the \(\mathrm{DE}\) in Problem 39 in Exercises \(2.1\). (c) If distance \(s\), measured from the point where the mass was released above ground, is related to velocity \(v\) by \(d s / d t=v(t)\), find an explicit expression for \(s(t)\) if \(s(0)=0\).

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

Electric Charge Field and Potential

Read Explanation

Energy Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Magnetism

Read Explanation

Fluids

Read Explanation

Turning Points in Physics

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.