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Chapter 17: Problem 62

The solution of the equation \(\frac{d y}{d x}=\frac{y}{2 y \log y+y-x}\) (A) \(x=y \log y+\frac{c}{y}\) (B) \(y=x \log x+\frac{c}{x}\) (C) \(x=-y \log y+\frac{c}{y}\) (D) None of these

Short Answer

Expert verified
(D) None of these

Step by step solution

01

Simplify the Differential Equation

Given the differential equation \(\frac{dy}{dx} = \frac{y}{2y \log y + y - x}\). First, simplify the right-hand side: \(2y \log y + y = y(2\log y + 1)\). The equation becomes \(\frac{dy}{dx} = \frac{y}{y(2\log y + 1) - x}\).
02

Change of Variables

Introduce a substitution to simplify the equation. Let \(v = \log y\), then \(y = e^v\), and consequently, \(\frac{dy}{dx} = y\frac{dv}{dx}\). Substitute into the equation, the equation becomes \(\frac{dv}{dx} = \frac{1}{2v + 1 - \frac{x}{e^v}}\).
03

Separate Variables

Separate the variables by expressing the equation as \(dx = e^v (2v + 1) dv\). Integrate both sides with respect to \(v\) and \(x\).
04

Integrate Both Sides

Integrate \(dx = e^v (2v + 1) dv\). The integral \(\int dx = x + C\), and the right side is integrated as \(\int e^v (2v + 1) dv\).
05

Solve the Integral

The integral \(\int e^v (2v + 1) dv\) can be approached using integration by parts. Let \(u = 2v + 1\) and \(dv = e^v dv\). The integral becomes \(e^v(2v + 1) - 2 \int e^v dv = e^v(2v + 1) - 2e^v = e^v(2v - 1)\).
06

Solve for x in terms of y

Substitute back \(v = \log y\), thus \(e^v = y\), to get \(x = y(2\log y - 1) + C\). Rearrange the equation to express in the form matching the choices given: \(x = 2y \log y - y + C\).
07

Match the Solution

Check the options given in the problem. The derived expression \(x = 2y \log y - y + C\) does not match any of the given choices directly.

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Key Concepts

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

Separation of Variables
The separation of variables technique is a foundational method for solving differential equations. It works by rearranging a differential equation so that each variable appears on a separate side of the equation. In the original problem, we encounter the equation \[ \frac{dv}{dx} = \frac{1}{2v + 1 - \frac{x}{e^v}}. \]To use separation of variables, we manipulate the equation to have all terms involving \(v\) and \(dv\) on one side, and all terms involving \(x\) and \(dx\) on another.
  • Rewriting: We obtain \( dx = e^v(2v + 1) dv \), where the left side is ready for integration with respect to \(x\), and the right side with respect to \(v\).
  • This approach simplifies the integration process by isolating each variable, making it much easier to solve the integral on each side.
Once segregated, each side of the equation can be integrated independently, greatly facilitating the solution of differential equations of this type.
Integration by Parts
Integration by parts is a technique for integrating products of functions, typically expressed in the form \( \int u \, dv = uv - \int v \, du \). In our exercise, integration by parts aids in solving the somewhat complex integral \[ \int e^v(2v + 1) \, dv. \]Here's how we apply it to this integral:
  • We select \( u = 2v + 1 \) and consequently, \( dv = e^v \, dv \).
  • Computing their derivatives (and antiderivatives) gives us \( du = 2 \, dv \) and \( v = e^v \).
  • Substituting these into the integration by parts formula yields \( e^v(2v + 1) - 2 \int e^v \, dv = e^v(2v - 1) \).
This process not only simplifies the integral but also allows us to find an analytic solution by breaking down complex expressions into manageable computations.
Substitution Method
Substitution in the context of differential equations helps transform a complicated expression into a simpler one. In this exercise, the substitution aims to simplify complex logarithmic terms.We make the substitution \( v = \log y \), then \( y = e^v \). This alters our differential equation from one involving \( y \) to one involving \( v \). The purpose here is to streamline the integration process.
  • This substitution gives us \( \frac{dy}{dx} = y \frac{dv}{dx} \), transforming the original equation into a format that's easier to manage.
  • The new variable \( v \) replaces the logarithmic component, simplifying as we manipulate the equation into \( \frac{dv}{dx} = \frac{1}{2v + 1 - \frac{x}{e^v}} \).
By strategically choosing a substitution, we change the differential equation into a form that's more straightforward to solve using other techniques like separation of variables or integration by parts, ultimately leading us to a solution more efficiently.

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Most popular questions from this chapter

Assertion: The order of the differential equation, of which \(x y=c e^{x}+b e^{-x}+x^{2}\) is a solution, is 2 . Reason: The differential equation is \(x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}-x y+x^{2}-2=0\)

The differential equation of all circles passing through the origin and having their centres on the \(x\)-axis is (A) \(x^{2}=y^{2}+x y \frac{d y}{d x}\) (B) \(x^{2}=y^{2}+3 x y \frac{d y}{d x}\) (C) \(y^{2}=x^{2}+2 x y \frac{d y}{d x}\) (D) \(y^{2}=x^{2}-2 x y \frac{d y}{d x}\)

The solution of the differential equation \(\frac{d y}{d x}=\frac{x+y}{x}\) satisfying the condition \(y(1)=1\) is (A) \(y=\ln x+x\) (B) \(y=x \ln x+x^{2}\) (C) \(y=x e^{(x-1)}\) (D) \(y=x \ln x+x\)

The solution of the equation \(\sin y \frac{d y}{d x}=\cos y(1-x\) \(\cos y\) ) is (A) \(\sec y=(1+x)+c e^{x}\) (B) \(\tan y=(1+x)+c e^{x}\) (C) \(\sec y=(1+x)+c e^{-x}\) (D) None of these

The solution of the differential equation \(\left(x^{2} y-2 x y^{2}\right) d x-\left(x^{3}-3 x^{2} y\right) d y=0\) is (A) \(\frac{x}{y}-2 \log x+3 \log y=c\) (B) \(\frac{x}{y}+2 \log x+3 \log y=c\) (C) \(\frac{x}{y}-2 \log x-3 \log y=c\) (D) None of these
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