Solving a Non-Exact ODE: $(y-2x^2y)dx +xdy = 0$

  • Context: MHB 
  • Thread starter Thread starter NotaMathPerson
  • Start date Start date
  • Tags Tags
    Ode
Click For Summary

Discussion Overview

The discussion revolves around solving the non-exact ordinary differential equation (ODE) given by $$(y-2x^2y)dx +xdy = 0$$. Participants explore methods for determining an integrating factor and discuss the separability of the equation, as well as the implications of using different techniques for solving it.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant asserts that the ODE is in exact form but fails the exactness test, leading them to use the technique $$\frac{M_y-N_x}{N}$$ to find an integrating factor, resulting in $u(x)=-2x$.
  • Another participant suggests that the ODE is separable and proposes factoring out the $y$ from the $dx$ term.
  • A later reply emphasizes the need to exponentiate the integral of $(M_y-N_x)/N$, proposing $u(x)=e^{-x^2}$ as a different integrating factor.
  • One participant agrees with the separability approach and provides a transformed version of the ODE after applying the integrating factor $e^{-x^2}$, claiming it becomes exact.

Areas of Agreement / Disagreement

Participants express differing views on the best approach to solve the ODE, with some advocating for the use of an integrating factor while others suggest treating it as separable. The discussion remains unresolved regarding the effectiveness of the initial method used by the first participant.

Contextual Notes

There are limitations in the assumptions made about the exactness of the ODE and the validity of the integrating factors proposed. The discussion does not resolve the mathematical steps involved in determining the correct integrating factor.

NotaMathPerson
Messages
82
Reaction score
0
Solve the ode

$$(y-2x^2y)dx +xdy = 0$$

The equation is in exact form $$Q(x,y)dx+ P(x,y)dy =0$$

When I test for exactness it fails. Then I used the technique $$\frac{M_y-N_x}{N}$$

I get

$u(x)=-2x$ as my integrating factor.

But I end still end up with a non-exact d.e why is that?

$$(-2xy+4x^3y)dx-2x^2dy=0$$
 
Last edited:
Physics news on Phys.org
It's separable. Factor out the $y$ on the $dx$ term.
 
Ackbach said:
It's separable. Factor out the $y$ on the $dx$ term.

Yes i did that already. What I am after is the answer as to why the method i used above did not work.
 
You need to exponentiate the integral of $(M_y-N_x)/N$. That is, in general,
$$u(x)=\exp\left[\int\left(\frac{M_y-N_x}{N}\right)dx\right].$$
I get $u(x)=e^{-x^2}$. See if you agree.
 
NotaMathPerson said:
Solve the ode

$$(y-2x^2y)dx +xdy = 0$$

The equation is in exact form $$Q(x,y)dx+ P(x,y)dy =0$$

When I test for exactness it fails. Then I used the technique $$\frac{M_y-N_x}{N}$$

I get

$u(x)=-2x$ as my integrating factor.

But I end still end up with a non-exact d.e why is that?

$$(-2xy+4x^3y)dx-2x^2dy=0$$

I agree with Ackbach that treating the given ODE as separable is the way to go here. However, if you wish to compute an integrating factor, you should use:

$$\mu(x)=\exp\left(\int -2x\,dx\right)=e^{-x^2}$$

And now, your ODE becomes:

$$\left(e^{-x^2}y\left(1-2x^2\right)\right)dx+\left(xe^{-x^2}\right)dy=0$$

And you can see by inspection this ODE is exact. :)
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K