Help Solving a System Of DEQ's


by ajohncock
Tags: solving
ajohncock
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#1
Mar23-09, 04:25 PM
P: 7
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 in to Differential Equations, so I am new at this stuff.
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de_brook
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#2
Mar23-09, 04:30 PM
P: 74
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
ajohncock
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#3
Mar23-09, 04:53 PM
P: 7
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).

tiny-tim
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#4
Mar23-09, 05:14 PM
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Help Solving a System Of DEQ's


Hi ajohncock! Welcome to PF!
Quote Quote by ajohncock View Post
(D+3)x + (D+3)y=2
erm can't you just solve this on its own?
HallsofIvy
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#5
Mar23-09, 05:22 PM
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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.
ajohncock
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#6
Mar23-09, 05:50 PM
P: 7
Quote Quote by HallsofIvy View Post
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.
tiny-tim
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#7
Mar23-09, 06:03 PM
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Quote Quote by ajohncock View Post
(D+3)x + (D+3)y=2
solution ?
ajohncock
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#8
Mar23-09, 06:05 PM
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Quote Quote by tiny-tim View Post
solution ?
Haha, I wish I knew what you meant by that.
tiny-tim
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#9
Mar23-09, 06:09 PM
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re-arrange it!
ajohncock
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#10
Mar23-09, 06:14 PM
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Quote Quote by tiny-tim View Post
re-arrange it!
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.
tiny-tim
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#11
Mar23-09, 06:19 PM
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Quote Quote by ajohncock View Post
I assume you mean to solve the equation for Dy
how is that re-arranging it?
ajohncock
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#12
Mar23-09, 06:22 PM
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Quote Quote by tiny-tim View Post
how is that re-arranging it?
Haha, maybe I'm not as math savvy as I thought. I guess I don't know what you mean by re-arrange it.
ajohncock
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#13
Mar23-09, 06:37 PM
P: 7
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.
tiny-tim
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#14
Mar24-09, 03:12 AM
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just got up

(D+3)(x + y) = 2 ?
de_brook
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#15
Mar24-09, 02:16 PM
P: 74
Quote Quote by ajohncock View Post
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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