Solving ODE: Integrating Factor for Problem 4d

Click For Summary
SUMMARY

The discussion focuses on solving the ordinary differential equation (ODE) represented by the equation (2x^2 + y)dx + [(x^2)y - x]dy = 0. The primary objective is to find an integrating factor to facilitate the solution. The user suggests utilizing the exponential function of the integral of a specific function with respect to x as the integrating factor. Additionally, the discussion emphasizes the importance of checking conditions on P and Q to determine the existence of roots and the necessity of grouping differentials to form exact differentials for integration.

PREREQUISITES
  • Understanding of ordinary differential equations (ODEs)
  • Familiarity with integrating factors in differential equations
  • Knowledge of exact differentials and their properties
  • Experience with homogeneous equations and their solutions
NEXT STEPS
  • Study the method of finding integrating factors for first-order ODEs
  • Learn about exact differential equations and how to identify them
  • Explore the process of grouping differentials to form exact differentials
  • Review techniques for solving homogeneous differential equations
USEFUL FOR

Mathematics students, educators, and anyone involved in solving ordinary differential equations, particularly those seeking to enhance their understanding of integrating factors and exact differentials.

roryhand
Messages
2
Reaction score
0
Howdy, I've read this forum for some time, however this is my first post. I am attempting to solve this ODE. I am looking to find an integrating factor, then solve. I have attached the link to the problem set if my input here is ambiguous. Number 4d. Thank you kindly for any help you might lend.


(2x^2)+y+((x^2)*y)-x)dy/dx=0

My reasoning takes me as far as the integrating factor being exp(int( ? )dx)

https://people.creighton.edu/%7Elwn70714/DE_Assignments/DE%20Assignment%202%20PDF.pdf
 
Last edited by a moderator:
Physics news on Phys.org
(2x^2)+y+((x^2)*y)-x)dy/dx=0

(2x^2+y)dx + [(x^2)*y-x]dy=0

Now you just need re-read your text-book about how to solve Pdx+Qdy=0, after checking some conditions on P&Q if such an equation has roots or none.
 
Or can approach it this way:

[tex](2x^2+y)dx+(x^2y-x)dy=0[/tex]

So, after a quick check for homogeneous, exact, and explicit calc. of an integrating factor via partials, we expand the differentials and attempt to group them together to form exact differentials:

[tex]2x^2dx+ydx+x^2ydy-xdy=0[/tex]

Well, the ydx-xdy can be grouped as:

[tex]y^2\left(\frac{ydx-xdy}{y^2}\right)[/tex]

This leaves us with:

[tex]2x^2dx+x^2ydy+y^2 d\left(\frac{x}{y}\right)[/tex]

Can you re-arrange this now to obtain exact differentials which can be integrated?
 

Similar threads

  • · Replies 25 ·
Replies
25
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
1
Views
3K
Replies
7
Views
3K
Replies
1
Views
2K
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 27 ·
Replies
27
Views
4K