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First order linear differential equation

  1. Nov 30, 2008 #1
    1. The problem statement, all variables and given/known data
    Solve this differential equation:
    (y^2 +1)*dx + (2xy + 1)*dy = 0

    2. Relevant equations
    dy/dx + P(x)*y = Q(x)
    u(x) = e^(integral of P(x)dx)
    (d/dx)(u(x)*y) = Q(x)*u(x)
    y = (integral of (Q(x)*u(x)dx))/(u(x)

    3. The attempt at a solution
    I tried dividing by dx then distributing and rearranging to get it into the right form, but run into problems:
    y^2 + 1 + 2xy*dy/dx + dy/dx = 0
    dy/dx + y/2x = (-1/(2xy))(dy/dx) -1/(2xy)

    this is the closest I could get it to the right form. It would give me u(x) = x^(1/2), but I wouldn't be able to integrate the right side as it would have both x and y. Is there a way to get past this, or did I just rearrange poorly? I just integrated it anyway and the part that was integrated with respect to x I held y as constant, and vice versa, but I'm sure it's wrong so I won't show how I did that. This is for a calc II class so it should be doable without any advanced tricks.

  2. jcsd
  3. Nov 30, 2008 #2


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    That's an exact differential. You should show us how you did it wrongly by integrating. Because you should be able to do it that way.
  4. Nov 30, 2008 #3
    that's interesting. We haven't done exact differentials as far as I know, but since I don't know what that means perhaps we have!

    dy/dx + y/2x = (-1/(2xy))(dy/dx) -1/(2xy)
    using u(x) = x^(1/2)

    integral of ((d/dx)(y*x^(1/2))dx) = integral of (-1/(2*x^(1/2)*y)(dy/dx)(dx)) - integral of (-1/(2*x^(1/2)*y)(dx))

    I integrated with respect to y for the first one on the right side, treating x as a constant, and integrated with respect to x on the right side, holding y constant, to get:

    x^(1/2)*y = -1/(2*x^(1/2))*ln(abs(y)) - (x^(1/2))/y + C
    y = -1/(2*x)*ln(abs(y)) - 1/y + C/(x^(1/2))
  5. Nov 30, 2008 #4


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    Another way to about doing this is to consider what d/dx( xy2) works out to be i.e. d(xy2)
  6. Dec 1, 2008 #5


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    Not quite. You are looking for a solution of the form f(x,y)=C. The differential of f is (df/dx)*dx+(df/dy)*dy (the derivatives are partial). So you are looking for a function such that df/dx=(y^2 +1) and df/dy=(2xy + 1). Like rock.freak667 points out, you could probably guess the answer.
  7. Dec 1, 2008 #6
    Oh, I think I know what to do then. I can flip my strategy around and solve for x instead:

    (y^2 +1)*dx + (2xy + 1)*dy = 0
    dividing by dy, distributing, and rearranging:
    dx/dy + 2x/y = - 1/y^2 - (1/y^2)*dx/dy

    therefore u(y) = e^(2*lny) = y^2


    (d/dy)((y^2)*x) = -1 - dx/dy
    integrating with respect to y
    (y^2)*x = -y - x + C

    Is this correct? Also, a question regarding my last integration with respect to y: when I integrate (dx/dy)*dy, I'm told that in situations like this "the dy's don't actually cancel out, but you can treat it as such." What *actually* happens then?

    Thanks for the help!
  8. Dec 2, 2008 #7


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    You've got the right answer. f(x,y)=x*y^2+x+y=C. So (df/dx)*dx+(df/dy)*dy=0. I'm a little confused by your last question. If you are integrating dy, then C could be any function of x. But if you differentiate d/dx and d/dy, you can figure out what it is, right?
  9. Dec 2, 2008 #8
    Oh, sorry, for my last question I meant regarding the step where I integrated

    integral of (-1-(dx/dy))dy

    I integrated by "distributing" the dy and "cancelling" it out on the right side to get

    int(-dy - dx)
    -y - x

    The step where I simplified (dx/dy)*dy to dx is what I was asking about. I remember hearing that technically that's not allowed, but it works out because it implies that you're actually doing something else, I think with substitution. I was wondering what implied substitution I was doing so I can more readily understand the integration (we've never worked with multiple variables before so it's a very new to me).
  10. Dec 2, 2008 #9


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    (dx/dy)*dy IS dx. Technically, it's not allowed to deal with 'infinitesimal' quantities at all, since they aren't 'real' numbers. But that's a technicality. Do it anyway. Everybody else does.
  11. Dec 3, 2008 #10

    This would probably a more organised method of arriving at the solution. What is also nice, is that you can test for the condition that it is an exact differential.
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