Results 1 to 5 of 5

Math Help - Homogeneous equation

  1. #1
    Member
    Joined
    Nov 2009
    Posts
    79

    Homogeneous equation

    Show that the given equation is homogeneous and solve the differential equation.

     dy/dx=x^2+3y^2/2xy<br />

    To show that it is homogeneous, I have dy/dx=1+3(y/x)^2/2(y/x)

    I am pretty sure that this is correct so far. Then,
    x(dv/dx)+v=1+3(y/x)^2/2(y/x)
    x(dv/dx)=1+3(y/x)^2/2(y/x)-v
    x(dv/dx)=(1+3v^2/2v)-v


    It is after this where I'm not quite sure what to do. According to my book the answer is x^2+y^2-cx^3=0. Thanks for any help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,368
    Thanks
    43
    Quote Originally Posted by steph3824 View Post
    Show that the given equation is homogeneous and solve the differential equation.

     dy/dx=x^2+3y^2/2xy<br />

    To show that it is homogeneous, I have dy/dx=1+3(y/x)^2/2(y/x)

    I am pretty sure that this is correct so far. Then,
    x(dv/dx)+v=1+3(y/x)^2/2(y/x)
    x(dv/dx)=1+3(y/x)^2/2(y/x)-v
    x(dv/dx)=(1+3v^2/2v)-v


    It is after this where I'm not quite sure what to do. According to my book the answer is x^2+y^2-cx^3=0. Thanks for any help
    It helps to use brackets. So

     <br />
\frac{dy}{dx} = \frac{x^2+3y^2}{2xy}<br />
    so if y = x v then \frac{dy}{dx} = x \frac{dv}{dx} + v

    so x \frac{dv}{dx} = \frac{1 + 3v^2}{2v} - v = \frac{1+v^2}{2v} as you've said. Now separate

     <br />
\frac{2v\,dv}{1+v^2} = \frac{dx}{x}<br />
and integrate.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2009
    Posts
    79
    Ok I integrated that and ended up with ln|1+v^2|=ln|x|+C

    Is that correct? Where do I go from here?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Calculus26's Avatar
    Joined
    Mar 2009
    From
    Florida
    Posts
    1,271
    exponentiate

    1+ v^2 = C x

    replace v = y/x

    simplify
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,368
    Thanks
    43
    Quote Originally Posted by steph3824 View Post
    Ok I integrated that and ended up with ln|1+v^2|=ln|x|+C

    Is that correct? Where do I go from here?
    Quote Originally Posted by Calculus26 View Post
    exponentiate

    1+ v^2 = C x

    replace v = y/x

    simplify
    Just an added note. As C is constant in ln|1+v^2|=ln|x|+C, replace it with \ln C and then follow Calc26.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. using the homogeneous equation...
    Posted in the Differential Equations Forum
    Replies: 6
    Last Post: February 16th 2011, 02:15 PM
  2. homogeneous equation
    Posted in the Differential Equations Forum
    Replies: 7
    Last Post: February 4th 2011, 04:19 PM
  3. [SOLVED] Is x^3 + y^3 -xy^2 dy/dx = 0 a homogeneous equation?
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: October 19th 2010, 08:03 PM
  4. Help with Homogeneous/Seperable equation
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: December 4th 2009, 08:51 PM
  5. non-homogeneous equation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 2nd 2007, 11:28 PM

Search Tags


/mathhelpforum @mathhelpforum