How to solve these Differential Equations?

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Discussion Overview

The discussion revolves around solving three differential equations, exploring methods such as exactness and integrating factors. Participants share their approaches and seek assistance with specific steps in the solution process.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Post 1 presents three differential equations and outlines initial approaches, including the use of homogeneous substitution and exactness.
  • Some participants suggest that the second equation is a nonhomogeneous linear ODE and that the homogeneous part is separable.
  • Post 3 claims that the first equation is exact and not homogeneous, and proposes an integrating factor for the second equation as simply y.
  • Post 4 expresses difficulty in finding the integrating factor for the first equation and shares a lengthy process for the third equation, ultimately finding an integrating factor of -tan(y) but struggles with its application.
  • Post 5 requests help with finding partial derivatives for the first and third equations, indicating a need for clarification on these steps.
  • Post 6 provides partial derivatives for the first equation, contributing to the discussion on exactness.

Areas of Agreement / Disagreement

Participants express differing views on the exactness of the equations and the effectiveness of proposed integrating factors. The discussion remains unresolved, with no consensus on the best approach for solving the equations.

Contextual Notes

Participants mention challenges with integrating factors and partial derivatives, indicating potential limitations in their methods and assumptions about the equations' properties.

Naeem
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Hi,

Q 1. ( e ^ y/x - y/x e ^ y/x + 1 / 1 + x^2 ) dx + e ^ y/x dy = 0

Ans. I know its Homogeneous sub, y = ux,
then, dy / dx = u dx + x du

I did this, and got to the point,

e^u dx + 1 / ( 1 + x^2) dx + x. e^u du = 0

How can we separate this now?

Q 2. y dx + ( 2x - y e^y ) dy = 0

I think we can use exactness, here

M = y

My = 1

N = 2x - y e^y

Nx = 2

Not exact,

so,

Integrating Factor would be : e ^ Integral My - Nx / N

Is this right. Integrating factor is getting to complicated to multiply the eqn, with. Any better way of doing this.


Q 3. ( 2x + tany) dx + ( x - x^2 tany ) dy = 0

I think here too, exactness, may be used, but any better way, if possible.
 
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The second one is really simple...

[tex]\frac{dx}{dy}=\frac{ye^{y}-2x}{y}[/tex]

Nonhomogenous linear ODE.The homogenous one is separable.

Daniel.
 
Alright, let's write these out:

1. [tex]\left(e^{\frac{y}{x}} - \frac{y}{x}e^{\frac{y}{x}} + \frac{1}{1+x^2}\right) \ dx + e^{\frac{y}{x}} \ dy = 0[/tex]

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

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

#1 is exact (check for yourself!). It isn't homogeneous.

An integrating factor that will work for #2 is just [itex]y[/itex] (you can determine it by inspection...).

Integrating factor isn't that bad for #3 (although you will definitely know which function to integrate after you find it...).
 
So, far had now luck with part 1,

If it is exact,

I tried to find partial derivatives with respect to y and x for both terms, and then applied

My - Nx / N

and the Integrating factor gets really messy...

I got 2 ,,, piece of cake.

3. if M = 2x + tany
then

My = sec^2y

N = x - x^2 tany

Nx = 1 -2xtany

Then,

Integrating factor = Nx - My / M

1-2xtany - sec^2y / 2x tany

Trig, identity,

sec^2y - tan^2y = 1
sec^2y = 1 + tan^2y

Plugging in this into: 1-2xtany - sec^2y / 2x tany , for sec^2y

we get,

1-2xtany -1 -tan^2y / 2x tany

After simplifying the above, I got:

Integrating factor as - tany

When I try to multiply both sides by -tany, the expression doesn't get any simpler, but infact becomes harder to evaluate,

Any ideas, on how this could be made simpler.

Thanks
 
1. How the heck to find the partial for no. 1 with respect to x and y...

It is a pain. I would appreciate if somebody could help me with number 1 and 3


Thanks
 
[tex]\left(e^{\frac{y}{x}} - \frac{y}{x}e^{\frac{y}{x}} + \frac{1}{1+x^2}\right) \ dx + e^{\frac{y}{x}} \ dy = 0[/tex]

I can't help you with solving the equation but the partials:

[tex]M_y = \frac{e^{\frac{y}{x}}}{x} - \frac{e^{\frac{y}{x}}}{x}+\frac{ye^{\frac{y}{x}}}{x} = \frac{ye^{\frac{y}{x}}}{x}[/tex]

[tex]N_x = \frac{-ye^{y/x}}{x^2}[/tex]
 

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