1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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?
    [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
    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


    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook