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Homework Help: 2 differentials, one of which is almost solved

  1. Jan 31, 2010 #1
    1. The problem statement, all variables and given/known data
    so, problem 1:
    xy' + 2y = 6x2y1/2

    2. Relevant equations
    so this is a bernoulli, where, in the form y' + p(x)y = q(x)yn

    3. The attempt at a solution
    xy' + 2y = 6x2y1/2
    y' + (2/x)y =6xy1/2
    so in the form, p(x) = 2/x, q(x) = 6x, and v = y1/2
    dividing both sides by y1/2
    y-1/2y' + (2/x)y1/2 = 6x

    substitute v = y1/2 and simplify
    this becomes:
    2dv/dx + (2/x)v = 6x
    dv/dx + (1/x)v = 3x
    this is linear, and let mu(x)= eint. 1/x dx = eln x = x
    multiply both sides, and get
    d/dx(xv) = 3x2
    xv = x3 + c0
    v = x2 + c0/x
    replace v
    y1/2 = x2 + co/x
    y = (x2 + c0/x)2

    here's the problem. when i do this without the constant of integration, it works fine, i.e. (x2)2 is a valid solution. so i've gone over this several times, but can't find my mistake. any help?

    problem no 2: i can't get started. a point in the right direction would be appreciated.

    ex+ y*exy + (ey+ x*eyx)y'=0

    as i type it, i wonder if i copied something down wrong, having the yx in reverse alphabetical order. oh well, yeah, a pointer would help. thanks!
  2. jcsd
  3. Jan 31, 2010 #2
    You did nothing wrong for the first problem. Just try plugging the terms into the original ODE again carefully.

    For the 2nd problem, make y' the subject and think about what happens when something is implicitly differentiated. I think the answer is ey + ex + exy = 0.

    Good luck.
  4. Jan 31, 2010 #3


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    Your answer to the first problem is correct. Perhaps your error occurs while trying to check your answer.

    In the second problem, you can rewrite it slightly:


    Does this form jog your memory?
  5. Jan 31, 2010 #4
    how about one more small hint?
  6. Jan 31, 2010 #5


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    It's an exact differential. You can see this because

    [tex]\frac{\partial}{\partial y}(e^x+ye^{xy}) = \frac{\partial}{\partial x}(e^y+xe^{xy})[/tex]

    So there's some function [itex]\Phi(x,y)[/itex] such that

    [tex]d\Phi = (e^x+ye^{xy})dx+(e^y+xe^{xy})dy[/tex]
  7. Jan 31, 2010 #6
    i seem to remember my last class skipping over this section, but i have the book and i'll see what i can do. thanks.
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