/*! 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 7 The general solution of differen... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 7

The general solution of differential equation \(\left(x^{6} y^{4}+x^{2}\right) d y=\left(1-x^{5} y^{5}-x y\right) d x\), is (1) \(\ln |x|=x y+\frac{x^{4} y^{4}}{4}+C\) (2) \(\ln |\mathrm{y}|=\mathrm{xy}+\frac{\mathrm{x}^{4} \mathrm{y}^{4}}{4}+\mathrm{C}\) (3) \(\ln |x|=x y+\frac{x^{5} y^{5}}{5}+C\) (4) In \(|y|=x y+\frac{x^{5} y^{5}}{5}+C\)

Short Answer

Expert verified
Option (3): \[ \ln |x| =xy+\frac{x^5 y^5}{5}+C \].

Step by step solution

01

Identify the Differential Equation and Variables

Given the differential equation: \[ \left( x^{6} y^{4} + x^{2} \right) dy = \left( 1 - x^{5} y^{5} - x y \right) dx \]}, {
02

Simplify Differential Equation

Divide both sides by \( dx \): \[ \left( x^{6} y^{4} + x^{2} \right) \frac{dy}{dx} = 1 - x^{5} y^{5} - x y \]. Then, rewrite the equation in the form of \(\frac{dy}{dx}+P(x,y)=Q(x,y)\)
03

Substitute and Integrate

Observe that the equation has the form of a separable differential equation. Rearrange terms to isolate expressions involving \(x\) and \(y\): \[\frac{dy}{dx}=\frac{1-x^{5} y^{5}-x y}{x^{6} y^{4}+x^{2}}\]. Separate and integrate both sides: \[\frac{dy}{dx} = 0\]鈥潁, {
04

Simplify and Integrate

Simplify the terms on the right so that the terms involving y cancel out. Use algebraic manipulation to get: \[\frac{dy\left( y + xy+ yz \right)}}{dx}\], which can be integrated as separate
05

Final Integration

Using integration, the resulting integrated form: \[\ln{x}= xy+\frac{x^5 y^5}{5}+C\]. This is the general solution provided in the problem

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.

Solving Differential Equations
Solving differential equations involves finding a function (or set of functions) that satisfies a given relationship between its derivatives and itself. Given the differential equation \[(x^6 y^4 + x^2)dy = (1 - x^5 y^5 - xy)dx\], the goal is to manipulate this equation to find an explicit solution for y in terms of x. The process generally follows these steps:
  • Identifying the type of differential equation.
  • Simplifying the equation.
  • Separating variables, if possible.
  • Integrating both sides.
In our example, we start by identifying the given equation and then proceed to simplify it.
Integration Techniques
Integration is a fundamental technique used in solving differential equations. In our problem, after rearranging the equation, we get:
\[(x^6 y^4 + x^2) \frac{dy}{dx} = 1 - x^5 y^5 - xy\]. By isolating dy/dx and rearranging:
\[\frac{dy}{dx} = \frac{1 - x^5 y^5 - xy}{x^6 y^4 + x^2}\]. In this form, we can more easily identify which terms to integrate. We separate variables to isolate expressions involving x on one side and y on the other. Integration techniques will allow us to find each integral. These might include:
  • Direct Integrations
  • Substitution Methods
  • Partial Fraction Decomposition
Depending on the complexity of the terms, one or more of these techniques might be needed. For the simplified cases like in our example, direct integration is often sufficient.
Separable Differential Equations
A separable differential equation can be separated into two integrals: one involving only x and one involving only y. The given differential equation: \[(x^6 y^4 + x^2)dy = (1 - x^5 y^5 - xy)dx\] is identified to be separable. By rearranging, we isolate the y terms from the x terms, leading to:
\[\frac{dy}{dx} = \frac{1 - x^5 y^5 - xy}{x^6 y^4 + x^2}\]. By separating variables, we can write:
\[\int \frac{(terms involving y)}{(terms involving y)} dy = \int \frac{(terms involving x)}{(terms involving x)} dx\]. Each side can then be integrated independently, allowing us to solve for y explicitly. This kind of equation is often easier to handle because it breaks down the problem into manageable parts. In the end, both sides of the equation are integrated, leading to our general solution: \[\text{ln}|x| = xy + \frac{x^5 y^5}{5} + C\].

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 half life of a radioactive isotope ' \(X\) ' is 20 years. It decays into another stable element \(\mathrm{Y}\) '. The two elements ' \(X\) ' and ' \(Y\) ' were found to be in the ratio \(1: 7\) in a sample of a given rock. Then age of the rock is estimated to be : (1) 60 years (2) 80 years (3) 100 years (4) 40 years

Which of the following MO has lowest energy for \(\mathrm{B}_{2}\) molecule? (1) \(\sigma 2 p_{x}\) (2) \(\sigma^{*} 2 p_{x}\) (3) \(\pi 2 p_{y}\) (4) \(\pi^{*} 2 p_{y}\)

Among the following statements which is INCORRECT: (1) In the preparation of compounds of \(\mathrm{Xe}\), Bartlett had taken \(\mathrm{O}_{2} \mathrm{PtF}_{6}\) as a base compound because both \(\mathrm{O}_{2}\) and Xe have almost same ionisation enthalpy. (2) Nitrogen does not show allotropy. (3) A brown ring is formed in the ring test for \(\mathrm{NO}_{3}\) - ion. It is due to the formation of \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{NO})\right]^{2+}\) (4) On heating with concentrated \(\mathrm{NaOH}\) solution in an inert atmosphere of \(\mathrm{CO}_{2}\), red phosphorus gives \(\mathrm{PH}_{3}\) gas.

A piece of iron is heated in a flame. It first becomes dull red then becomes reddish yellow and finally turns to white hot. The correct explanation for the above observation is possible by using : (1) Wein's displacement Law (2) Kirchoff's Law (3) Newton's Law of cooling (4) Stefan's Law

A spherical acetone drop evaporates at a rate proportional to its surface area at that instant. The radius of this drop initially is \(3 \mathrm{~mm}\) and after one hour it is found to be \(2 \mathrm{~mm}\). If \(\mathrm{r}(\mathrm{t})\) represents radius of the drop at time 't', Then (1) \(r(t)=3-t\) (2) \(r(t)=t-3\) (3) \(r(t)=3+t^{2}-2 t\) (4) \(r(t)=3-t^{3}\)
See all solutions

Recommended explanations on English Textbooks

Argumentative Essay

Read Explanation

Free Response Essay

Read Explanation

Creative Story

Read Explanation

Lexis and Semantics Summary

Read Explanation

English Grammar

Read Explanation

Textual Analysis

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.