How to Isolate Variables in a System of Differential Equations?

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Discussion Overview

The discussion revolves around methods for isolating variables in a system of differential equations. Participants explore various approaches to manipulate the equations to facilitate solving for the unknown functions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses difficulty in isolating variables in the given system of equations and seeks assistance.
  • Another suggests substituting one equation into the other to reduce the system to a single second-order differential equation.
  • A participant attempts to manipulate the equations but struggles with terms that remain after substitution, particularly involving the variable y.
  • Some participants propose various methods, including multiplying equations by certain operators to eliminate variables, but face challenges with the resulting expressions.
  • There are multiple suggestions to rearrange the equations, but participants express confusion about the implications of these rearrangements.
  • One participant mentions the possibility of obtaining an integro-differential equation through manipulation, indicating a shift in the type of equation being dealt with.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method to isolate variables, and multiple competing approaches are presented without resolution.

Contextual Notes

Participants express uncertainty about how to handle specific terms that arise during their manipulations, indicating a lack of clarity in the steps needed to isolate variables effectively.

ajohncock
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Hi Guys,

I'm tying to solve a system of equations. I know I need to operate on the top and the bottom both in order to isolate the X's and Y's, but I can't seem to figure what to operate on them with. Here are the equations, any help is appreciated. Thanks

D2x - Dy=t
(D+3)x + (D+3)y=2

I should be able to finish solving it if I can just get them in the forms I need.

Edit: I have a feeling this is going to seem really obvious and easy when I see it. But I am just getting into Differential Equations, so I am new at this stuff.
 
Last edited:
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what about if you substitute Dy from (1) into (2), then you have a second order diffential equation in form of x and t. I believe x and y are function of t
 
If I do that then I get D2x + Dx + 3x - t + 3y = 2

Then I don't know how to deal with the -t + 3y

Thanks for the idea though. I hadn't thought of it.
Edit: After another look you could then write it as D2x + Dx + 3x = t - 3y + 2

But I still don't know what to do with it from here. I can solve the associated homogeneous equation, but then I still don't know what to do with the y on the right side. I think I have to get rid of either y or x entirely before I can solve for x(t) or y(t).
 
Welcome to PF!

Hi ajohncock! Welcome to PF! :smile:
ajohncock said:
(D+3)x + (D+3)y=2

erm :redface: … can't you just solve this on its own? :wink:
 
tiny-tim, are you asserting that he should be able to solve a single equation in two unknowns?

D2x - Dy=t
(D+3)x + (D+3)y=2

If you "multiply" the first equation by D-3 and the second equation by D you get
D2(D-3)x- D(D-3)y= (D-3)t= -3t
D(D+3)x+ D(D-3)y= 0 and then adding eliminates y:

D2(D-3)x+ D(D-3)x= -3t or D(D+1)(D-3)x= -3t.

Solve that equation for x, then put that x into D2x- Dy= t and solve that equation for y.
 
HallsofIvy said:
If you "multiply" the first equation by D-3 and the second equation by D you get
D2(D-3)x- D(D-3)y= (D-3)t= -3t
D(D+3)x+ D(D-3)y= 0 and then adding eliminates y:

D2(D-3)x+ D(D-3)x= -3t or D(D+1)(D-3)x= -3t.

Solve that equation for x, then put that x into D2x- Dy= t and solve that equation for y.

The problem with this is that, the part I BOLDED is actually (D+3) in the original equation. However that may be on the right track. I tried multiplying the top by (D+3) and the bottom by D.

I then added the equations and come up with this:

D2x(D+3) + D(D+3)x = 2D + 3t + tD

Simplified that is D3x + 4D2x + 3Dx = 2D + 3t + Dt

Which does get rid of the y, but now I am unsure what to do with the right side.

This is frustrating.
 
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ajohncock said:
(D+3)x + (D+3)y=2

solution … ? :smile:
 
tiny-tim said:
solution … ? :smile:

Haha, I wish I knew what you meant by that.
 
re-arrange it! :wink:
 
  • #10
tiny-tim said:
re-arrange it! :wink:

I assume you mean to solve the equation for Dy and substitute that back into the other equation. But when I solve it for Dy I get:

Dy = -Dx - 3x - 3y +2 Which is all fine and good except for the -3y. Which poses a problem when substituted back into the first equation.
 
  • #11
ajohncock said:
I assume you mean to solve the equation for Dy …

how is that re-arranging it? :confused:
 
  • #12
tiny-tim said:
how is that re-arranging it? :confused:

Haha, maybe I'm not as math savvy as I thought. I guess I don't know what you mean by re-arrange it.
 
  • #13
Man I've been looking at this and manipulating it for too long. I've got nothing! It's the right side I don't know how to deal with.
 
  • #14
just got up :zzz: …

(D+3)(x + y) = 2 ? :wink:
 
  • #15
ajohncock said:
If I do that then I get D2x + Dx + 3x - t + 3y = 2

Then I don't know how to deal with the -t + 3y
since y is a function of t and x then if Dy = (D^2)x -t, you can apply integration so that you now obtain what is called integro-differential equation in form of x and t
 

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