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Separable ODE

  1. Sep 30, 2015 #1
    1. The problem statement, all variables and given/known data
    Find a one-parameter family of solutions to the isobaric equation

    (x+y)dx - (x-y)dy = 0

    3. The attempt at a solution
    First I subtracted -(x-y) to the both sides
    Step 1: (x+y)dx = (x-y dy

    I then distributed dx into (x+y) and dy into (x-y) to get
    Step 2: xdx +ydx = xdy -ydy

    I then divided the y on the left side of the equal sign to both sides
    Step 3: x/ydx + dx = x/ydy + dy

    Step 4: using 1/u = x/y

    I get
    Step 5: 1/u dx + dx = 1/u dy - dy

    I then factored out dx and dy to get
    Step 6: (1/u +1)dx = (1/u - 1)dy

    I then divided (1/u - 1) to both sides and dx to both sides to get
    Step 7: (1/u +1) / (1/u - 1) = dy/dx

    using the above 1/u = x/y and rearranging to the form y = xu and taking derivative of y in respect to x I get
    Step 8: dy/dx = (du/dx)x + u

    I then used dy/dx in step 8 to plug into step 7 dy/dx to get
    Step 9: (1/u +1) / (1/u - 1) = (du/dx)x + u

    I then subtracted the u on the right side of the equal sign to both sides to get
    Step 10: (-u+u^3) / 1 - u = (du/dx)x

    I then used separable equations to get
    Step 11: (1/x)dx = (1-u) / (-u + u^3)du

    This is where I stopped because I don't like (u cubed on the denominator).

    Any help would be fantastic. I know my next steps will be integration but I feel there's another step that needs to be taken.
  2. jcsd
  3. Oct 1, 2015 #2


    Staff: Mentor

    Or u = y/x, which is a lot simpler. Use this substitution to get a separable differential equation.
  4. Oct 1, 2015 #3

    Thank you, Mark.

    I'm going to work the problem from the beginning so I can use u = y/x.

  5. Oct 2, 2015 #4
    Hi Mark,

    Thank you again. I successfully finished the problem. My answer determined was arctan(1 + y2/x2) - 1/2ln|1 + y2/x2| = ln|x| + c
  6. Oct 5, 2015 #5
    Last edited: Oct 5, 2015
  7. Oct 5, 2015 #6
    can someone explain to me how to post latex?
    [tex]\int x^2 dx [/tex]
  8. Oct 5, 2015 #7


    Staff: Mentor

    I didn't check the OP's answer. When you solve a DE, it's always a good idea to check that your solution satisifies the original diff. equation.
    Does the above have anything to do with this problem?

    See https://www.physicsforums.com/help/latexhelp/ under INFO --> Help/How-To, in the banner at the top of the page.[/QUOTE]
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