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What am I doing wrong? (D.E.)

  1. Jun 22, 2005 #1
    I have solved two Differential Equations; my answers are very similar to the provided general answers, but I just cannot get to them. Would someone tell me what I was doing wrong in my process?
    1.
    [x^2-2y^2]dx + xy dy = 0
    xy dy = [2y^2 - x^2]dx
    dy/dx = 2y/x - x/y
    dy/dx - 2/xy = -x(y^-1)
    Multiply by y
    y(dy/dx) - (2/x) y^2 = -x
    Let v= y^2
    Then
    dv/dx - 4 (x^-1) V = -2x
    Since p(x) = -4/x, e ^integrate -(x) = x^-4
    Multiply by x^-4
    we have d (x^-4 v) /dy = -2x * x^-4 = -2(x^-3)
    Integrate -2(x^-3)
    we have x^-2 + C
    Hence x^-4 * v = x^-2 + C
    since v=y^2,
    x^4*y^2 = x^-2 + C
    y^2 = x^2 + C(x^4)
    y^2-x^2 = C(x^4)
    x^4 = C^-1 (y^2 - x^2)
    But the general answer is x^4 = C(y^2 - x^2). What did I do wrong?

    2. y dx + [x^2 - x] dy = 0
    y dx = [x - x^2] dy
    dx/dy = x/y - x^2/y
    dx/dy - x/y = -(x^2/y)
    Multiply by x^-2
    x^-2 (dx/dy) - y^-1 * x^-1 = - (y^-1)
    Let v = x^-1 then dv/dx = -(x^-2)(dx/dy)
    then dx/dy + y^-1*v = y^-1
    Then p(y) = y^-1
    Calculate e^integrate p(y) we have y
    so multipl by y
    y (dv/dy) = y* y^-1* v = y^-1 * y
    d (y*v)/dy = 1
    integrate 1 and we have
    y*v = y + C
    since v = x^-1
    y*x^-1 = y + C
    y = yx + Cx
    y-yx = Cx
    y(1-x) = Cx But the general answer is y(x+1) = Cx

    Please trust me I tried everything I could think of to fix the problems, but I couldn't. Every time I redo the problems, I got the same answers. With my knowledge of differential equation (I have just started 2 weeks ago), I am out of ideas).

    I also have posted a question regarding different problem which I could not solve. I would appreciate it if you take a look at that question and instruct me how I should solve them (some people tried to help me but I still cannot get it). Right now I don't know either how to obtain IF from f(xy)ydx + f(xy)x dy = 0 equations nor change the form to dy/dx + p(x)y = c
    Thank you.
     
  2. jcsd
  3. Jun 22, 2005 #2

    HMS

    User Avatar

    Agreed, you have taken the term 2y/x to the left, but why did it become 2/xy? May be you can start correcting from here. Hope that helps.
     
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