Results 1 to 6 of 6

Math Help - Homogeneous 1st order Differential Equation

  1. #1
    Newbie
    Joined
    Nov 2006
    Posts
    17

    Homogeneous 1st order Differential Equation

    Hi guys, I've gotten so far with this but I'm stuck on what to do next:

    Have I done the steps correctly? I followed what I was told but maybe I've made a mistake I cannot see.

    (2x^2+y^2)y'=2xy

    y'= \frac{2(y/x)}{2+(y/x)^2}

    So by letting v=y/x and differentiating and substituting my value for v and dy/dx in I get the following seperable DE

    \int\frac{(-2+v^2)}{v^3} dv = \int(1/x) dx

    Which I work out to be

    \frac{1}{v^2} -\ln v = \ln x +C

    But Im unsure how to isolate v, I tried using maple and It gave me an answer using the Lambert W function which I've never seen before, so I'm sure I've made a mistake along the lines..

    Any help is appreciated!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by dtwazere View Post
    Hi guys, I've gotten so far with this but I'm stuck on what to do next:

    Have I done the steps correctly? I followed what I was told but maybe I've made a mistake I cannot see.

    (2x^2+y^2)y'=2xy

    y'= \frac{2(y/x)}{2+(y/x)^2}

    So by letting v=y/x and differentiating and substituting my value for v and dy/dx in I get the following seperable DE

    \int\frac{(-2+v^2)}{v^3} dv = \int(1/x) dx

    Which I work out to be

    \frac{1}{v^2} -\ln v = \ln x +C

    But Im unsure how to isolate v, I tried using maple and It gave me an answer using the Lambert W function which I've never seen before, so I'm sure I've made a mistake along the lines..

    Any help is appreciated!
    Why not just back-substitute v = y/x and use a basic log rule to simplify ....? By the way, I think the -2 should be just 2.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Hmm. Standard form (at least how I learned it) is

    -2xy\,dx+(2x^{2}+y^{2})\,dy=0.

    Substitute x=uy, with dx=u\,dy+y\,du, giving us

    -2uy^{2}(u\,dy+y\,du)+(2u^{2}y^{2}+y^{2})\,dy=0, or

    -2uy\,du+dy=0. Separating yields

    -2u\,du+\dfrac{dy}{y}=0.

    Can you finish? I was able to solve for x, but not for y.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,408
    Thanks
    1294
    \displaystyle (2x^2 + y^2)\,\frac{dy}{dx} = 2x\,y

    \displaystyle \frac{2x^2 + y^2}{2x\,y} = \frac{dx}{dy}

    \displaystyle \frac{x}{y} + \frac{y}{2x} = \frac{dx}{dy}

    \displaystyle \frac{x}{y} + \frac{1}{2}\left(\frac{x}{y}\right)^{-1} = \frac{dx}{dy}.

    Now make the substitution \displaystyle v = \frac{x}{y} \implies x = v\,y \implies \frac{dx}{dy} = y\,\frac{dv}{dy} + v and the integral becomes

    \displaystyle v + \frac{1}{2}v^{-1} = y\,\frac{dv}{dy} + v

    \displaystyle \frac{1}{2}v^{-1} = y\,\frac{dv}{dy}

    \displaystyle \frac{1}{2}y^{-1} = v\,\frac{dv}{dy}

    \displaystyle \int{\frac{1}{2}y^{-1}\,dy} = \int{v\,\frac{dv}{dy}\,dy}

    \displaystyle \frac{1}{2}\ln{|y|} + C_1 = \int{v\,dv}

    \displaystyle \frac{1}{2}\ln{|y|} + C_1 = \frac{1}{2}v^2 + C_2

    \displaystyle v^2 = \ln{|y|} + C where \displaystyle C = 2(C_1 - C_2)

    \displaystyle \left(\frac{x}{y}\right)^2 = \ln{|y|} + C

    \displaystyle \frac{x^2}{y^2} = \ln{|y|} + C

    \displaystyle x^2 = y^2(\ln{|y|} + C)

    \displaystyle x = \pm y \sqrt{\ln{|y|} + C }.


    You will not be able to isolate \displaystyle y.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Nov 2006
    Posts
    17

    implicit

    Thanks for the help!!

    I forgot to add it says in the question to find the implicit solution,

    Would I be able to isolate it for x and then set it equal to a constant?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Quote Originally Posted by dtwazere View Post
    Thanks for the help!!

    I forgot to add it says in the question to find the implicit solution,

    Would I be able to isolate it for x and then set it equal to a constant?
    What is "it" that you're setting equal to a constant? And why would you want to set it equal to a constant? Neither x nor y is, in this problem, in general, equal to a constant.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. First Order Differential Equation (Homogeneous?)
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: March 6th 2011, 03:41 AM
  2. Second order non homogeneous differential equation
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: February 10th 2011, 02:21 AM
  3. Replies: 1
    Last Post: October 16th 2010, 12:19 PM
  4. Second Order Homogeneous Equation IVP
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: September 15th 2010, 12:38 AM
  5. second order differential constant coefficients homogeneous
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: October 1st 2008, 09:50 AM

Search Tags


/mathhelpforum @mathhelpforum