Help Solving a System Of DEQ's

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In summary: If you "multiply" the first equation by D-3 and the second equation by D you getD2(D-3)x- D(D-3)y= (D-3)t= -3tD(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
  • #1
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.
 
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  • #2
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
 
  • #3
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).
 
  • #4
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:
 
  • #5
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.
 
  • #6
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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  • #7
ajohncock said:
(D+3)x + (D+3)y=2

solution … ? :smile:
 
  • #8
tiny-tim said:
solution … ? :smile:

Haha, I wish I knew what you meant by that.
 
  • #9
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
 

What is a system of DEQ's?

A system of DEQ's (differential equations) is a set of equations that describe the relationship between the rate of change of one or more variables with respect to one or more independent variables. These equations are commonly used in mathematical modeling to describe the behavior of complex systems.

How do I solve a system of DEQ's?

Solving a system of DEQ's involves finding the values of the unknown variables that satisfy all of the equations in the system. This can be done through various methods such as substitution, elimination, or using matrices. It is important to note that there is not always a single solution to a system of DEQ's.

What are some real-world applications of using systems of DEQ's?

Systems of DEQ's have many real-world applications, including predicting population growth, modeling chemical reactions, and analyzing the spread of diseases. They are also used in engineering to design and optimize systems such as electrical circuits and mechanical systems.

What are the common challenges in solving a system of DEQ's?

One of the main challenges in solving a system of DEQ's is the complexity of the equations and the number of variables involved. This can make it difficult to find a solution, and in some cases, there may not be a closed-form solution. Another challenge is ensuring the accuracy of the solution, as small errors in the initial conditions or in the solving process can lead to significant differences in the final result.

Are there any software or tools available to help solve systems of DEQ's?

Yes, there are many software programs and online tools available to help solve systems of DEQ's. These include programs such as Mathematica, MATLAB, and Wolfram Alpha, as well as online calculators and equation solvers. It is important to double-check and verify the results obtained from these tools, as they may not always be accurate.

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