How can I improve my understanding of solving differential equations?

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Homework Help Overview

The discussion revolves around solving a specific differential equation of the form [y(y^3 - x)]dx + [x (y^3 + x)]dy = 0. Participants are exploring methods to transform this equation into a solvable format, specifically the dy/dx + P(x)y = Q(x) form, while also discussing the nature of the equation and its characteristics.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to identify the correct form of the differential equation and discussing potential methods such as using integrating factors and checking for exactness. Questions about the validity of their approaches and the need for integrating factors are raised.

Discussion Status

Some participants have offered alternative methods and insights into the nature of the equation, while others express uncertainty about their understanding and seek further clarification. There is an ongoing exploration of different approaches without a clear consensus on the best method to proceed.

Contextual Notes

One participant notes their recent start in a differential equations class and expresses difficulty in applying concepts to solve problems, highlighting a potential gap in foundational understanding.

Beez
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Hello, I have tried to solve the following problem but did not succeed to do so.

[y(y^3 - x)]dx + [x (y^3 + x)]dy = 0
I sense that the key factor here is (y^3 - x ) and (y^3 +x), but could not figure out how to lead the equation to

dy/dx + P(x)y = Q(x) form.

The general answer for the problem is 2xy^3 - x^2 = Cy^2.

Once I can change the equation to dy/dx + P(x)y = Q(x) form, I can do the rest (probably anybody can...)

Thanks for your help in advance.
 
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This is of the form f(xy)ydx +F(xy)xdy=0
here the Integrating factor is ::
[tex]\frac{1}{xy[f(xy)-F(xy)]}[/tex]
and general Integral equation is ::
[tex]\int \frac{f(xy)+F(xy)}{f(xy)-F(xy)} \frac{d(xy)}{xy} + log\frac{x}{y} = c[/tex]
 
Beez said:
[y(y3 - x)]dx + [x (y3 + x)]dy = 0
that looks awfully a lot like exact DE: you'll have to get partial derivatives and the whole nine yards. Try that.
 
Well , there is an alternative if you use differentials
now
d(xy)= ydx +xdy
and
[tex]d(\frac{y}{x}) = \frac{xdy-ydx}{x^2}[/tex]

Here u can rearrange ur diff eqn to

:: y3 d(xy) + x3d(y/x)=0
divide by y3 u will get the required answer after integrating
 
Exact D.E.

I believe it is an exact DE since I have just learned that part. But when I did
\partial M (x, y) / \partial y = 4y^3 - x and
\partial N (x, y) / \partial x = y^3 + 2x so they are not the same.

But I could not find integrating facutor to make their answers equal. What should I do now?
 
I will try that

Thank you "himanshu121". I will try that for now to see if I can understand that formula.


Well, I couldn't get it.

I have just started this differential equation class (independent). I thought I understood them well. However, when it comes to solve problems, I am experiencing a hard time. For example, this equation, I could not see why it Have some suggestion to improve my understanding?
 
Last edited:

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