# Help Solving a System Of DEQ's

#### ajohncock

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

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

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

Homework Helper
Welcome to PF!

Hi ajohncock! Welcome to PF!
(D+3)x + (D+3)y=2
erm … can't you just solve this on its own?

#### HallsofIvy

Homework Helper
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

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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Homework Helper

Homework Helper
re-arrange it!

#### ajohncock

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

Homework Helper
I assume you mean to solve the equation for Dy …
how is that re-arranging it?

#### ajohncock

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

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

Homework Helper
just got up :zzz: …

(D+3)(x + y) = 2 ?

#### de_brook

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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