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Differential Equations - Find equation of line

  1. Jun 27, 2010 #1
    1. The problem statement, all variables and given/known data

    A curve passing through (3,-2) has a slope given by (x^2 + y^2)/(y^3 - 2xy). Find the equation of the curve.

    2. Relevant equations

    3. The attempt at a solution

    My first thought was to plug in the points (3,-2) into the slope equation and plug them into the line equation (y - y1) = m(x - x1). Is that wrong? Seemed too easy for it to be a problem for differential equations..

    So, I found that i could make the slope equation into an exact equation and solved for it.
    I found that the answer i get is:
    with c = -17
    f(x,y) = x^3/3 + xy^2 - y^4/4 - 17
    However that is not the equation of the curve. Is that the equation of the tangent line at (3,-2)?
    I am stuck at this point, I am pretty sure I am done with 90% of the work, but I can't seem to figure out the equation of the curve from here on.
  2. jcsd
  3. Jun 27, 2010 #2


    Staff: Mentor

    Yes, you are pretty close, but you faltered a bit at the end.

    The equation you were solving looked something like this:
    Fx dx + Fy dy, where Fx = x2 + y2 and Fy = -y3 + 2xy

    The solution to the equation above with the partials is F(x, y) = C, so you should have gotten (1/3)x3 - (1/4)y4 + xy2 = C.

    Substitute y(3) = -2 into the equation above to find the constant C.
  4. Jun 27, 2010 #3
    Yes I did that, realized that C = 17 (not negative). But I am still puzzled, what is the equation of the curve?
  5. Jun 28, 2010 #4


    Staff: Mentor

    (1/3)x3 - (1/4)y4 + xy2 = 17.
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