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System of Differential Equations

  • #1

Homework Statement


dx/dt + dy/dt = 2x + 2y + 1

dx/dt + 2(dy/dt) = y + 3

Homework Equations



The Attempt at a Solution


Dx + Dy = 2x + 2y + 1

Dx + 2Dy = y + 3

--Rearranging----------------------------
Dx - 2x = -Dy + 2y + 1
Dx = -2Dy + y + 3
---Factoring-----------------------------
(D - 2)x = (-D + 2)y + 1
Dx = (-2D + 1)y + 3
--Eliminating x--
(D)(Dx-2x) = ((-D+2)y + 1) (D) <--- multiply by D
-(D-2)(Dx) = ((-2D + 1)y + 3) -(D-2) <---- multiply by -(D-2)

I get to this point and try to cancel out terms but it becomes a mess and attempting to find the solution usually leaves my with polynomials I can't factor, I don't know what to do..
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


dx/dt + dy/dt = 2x + 2y + 1

dx/dt + 2(dy/dt) = y + 3

Homework Equations



The Attempt at a Solution


Dx + Dy = 2x + 2y + 1

Dx + 2Dy = y + 3

--Rearranging----------------------------
Dx - 2x = -Dy + 2y + 1
Dx = -2Dy + y + 3
---Factoring-----------------------------
(D - 2)x = (-D + 2)y + 1
Dx = (-2D + 1)y + 3
--Eliminating x--
(D)(Dx-2x) = ((-D+2)y + 1) (D) <--- multiply by D
-(D-2)(Dx) = ((-2D + 1)y + 3) -(D-2) <---- multiply by -(D-2)

I get to this point and try to cancel out terms but it becomes a mess and attempting to find the solution usually leaves my with polynomials I can't factor, I don't know what to do..
Why not do it the easy way? Solve for Dx and Dy in terms of x and y, so you have a standard linear system of the form
[tex] \pmatrix{x'(t)\\y'(t)} = \pmatrix{a_1 & b_1 \\ a_2 & b_2} \pmatrix{x\\y} + \pmatrix{c_1\\c_2},[/tex]
where the ##a_i, b_i, c_i## are constants. Then use standard methods.
 
  • #3
I have very little Linear Algebra knowledge, I don't know if i could set up the
linear system correctly
|Dx| = |-Dy + 2x + 2y|
|Dy| = |-2Dy + y + 3 |
 
  • #4
Ray Vickson
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I have very little Linear Algebra knowledge, I don't know if i could set up the
linear system correctly
|Dx| = |-Dy + 2x + 2y|
|Dy| = |-2Dy + y + 3 |
I don't understand what you are writing here. Anyway, you do not need to know any linear algebra; you just have two linear equations in the two unknowns Dx and Dy, and you can solve them the way you learned back in school.
 
  • #5
I guess i'm confused because I have, derivatives on both sides when solving for Dx and Dy
 
  • #6
vela
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Here are the original differential equations:

dx/dt + dy/dt = 2x + 2y + 1
dx/dt + 2(dy/dt) = y + 3

Try subtracting the first equation from the second.
 

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