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Differential Equations, Subsitution Methods

  1. Sep 16, 2009 #1
    1. The problem statement, all variables and given/known data

    1. x(x+y)y' = y(x-y)

    2.y' = (4x + y)^2

    3. (x+y)y' = 1


    2. Relevant equations

    I have tried using the substitution method v =, the homogeneous method as well.

    3. The attempt at a solution

    For 1. i factored out the xy and got 1/xy y' = integral ( (1-y/x)/(1+(x/y)) but cant take it further to the final answer ln(xy) = (x/y) + C

    For 2. i tried sub v = 4x+y, but didnt even come close to the answer

    For 3. I tried homogenous equations making it (1+y/x)y' = 1/x, but i get stuck with 1/x^2 later on which i cant factor to make it a simple first order linear diff...

    PLEASE HELP!!
     
  2. jcsd
  3. Sep 17, 2009 #2

    Mark44

    Staff: Mentor

    How did you factor out xy? I don't see that xy is a factor, so I'm puzzled about how you were able to factor it out.
    Well, I tried the same substitution in #2 and was able to get a separable differential equation. Show me what you tried so I can see where you went wrong.
    Why not try a substitution u = x + y? If you do it correctly, you should get a separable differential equation.
     
  4. Sep 17, 2009 #3
    ok so i redid problems 2 & 3 and they worked for me...

    as for problem 1...

    I can factor down to y' = (y/x) ((1-y/x)/(1+y/x))

    but then i get the left side after using v = y/x to be (1/2)(1/v - lnv) = lnx + c but then i cant reduce it down to the answer in the back that is: ln(xy) = C + x/y....

    I did: y' = y(x-y)/x(x+y)

    then: y'/y = (x-y)/(x(x+y))

    then: factor out x from top of right side... y'/y = x(1-y/x)/(x(x+y)), so x's cancel

    i get: y'/y = (1-y/x)/(x+y) ok so then factor out x from bottom: y'/y = (1-y/x)/(x(1+y/x), then multiply y back....

    so: y' = (y/x)((1-y/x)/(1+y/x) let me know if i made a mistake here
     
  5. Sep 18, 2009 #4

    Mark44

    Staff: Mentor

    So far, so good.
    Now multiply both sides by 2 to get
    1/v - ln v = 2lnx + 2c = ln x2 + 2c
    so 1/v = ln v + ln x2 + 2c = ln[v x2] + 2c
    so x/y = ln(y/x * x2) + 2c = ln(xy) + C (where C = 2c)
    That's essentially what your book shows for the answer. The answer also seems to assume that x > 0 and y > 0, otherwise you would have some absolute values after integrating and getting the ln terms.

     
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