Results 1 to 6 of 6

Math Help - Seperable Differential equation

  1. #1
    Newbie
    Joined
    Apr 2010
    Posts
    11

    Seperable Differential equation

    Hey, I got stuck on the following differential equation. Given y = 0 when x = 2, solve \dfrac{x}{y}\cdot\dfrac{dx}{dy} = \dfrac{x^2+1}{y^2+1}.

    I can't separate the variables and I couldn't turn it into a form that I can take an integrating factor.
    Last edited by mr fantastic; June 9th 2010 at 07:12 PM. Reason: Edited title.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Deadstar's Avatar
    Joined
    Oct 2007
    Posts
    722
    Ok I'll separate the variables for you...

    \dfrac{x}{y}\cdot\dfrac{dx}{dy} = \dfrac{x^2+1}{y^2+1}

    => \dfrac{x}{x^2 + 1}dx = \dfrac{y}{y^2+1} dy


    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2010
    Posts
    11
    Quote Originally Posted by Deadstar View Post
    Ok I'll separate the variables for you...

    \dfrac{x}{y}\cdot\dfrac{dx}{dy} = \dfrac{x^2+1}{y^2+1}

    => \dfrac{x}{x^2 + 1}dx = \dfrac{y}{y^2+1} dy


    So \int\dfrac{x}{x^2 + 1}dx = \int\dfrac{y}{y^2+1} dy

    \therefore \;\;\; \dfrac{1}{2}\log(x^2+1) = \dfrac{1}{2}\log(y^2+1)+C

    when y = 0, x = 2

    \therefore \;\;\; \dfrac{1}{2}\log(3) = C

    \therefore \;\;\; \dfrac{1}{2}\log(x^2+1) = \dfrac{1}{2}\log(y^2+1)+\log\sqrt{3}.<br />

    The question said in the begening express teh solution as an implicit equation relating x and y. That's the way it already is, right?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1
    Quote Originally Posted by Hibachi View Post
    So \int\dfrac{x}{x^2 + 1}dx = \int\dfrac{y}{y^2+1} dy

    \therefore \;\;\; \dfrac{1}{2}\log(x^2+1) = \dfrac{1}{2}\log(y^2+1)+C

    when y = 0, x = 2

    \therefore \;\;\; \dfrac{1}{2}\log(3) = C

    \therefore \;\;\; \dfrac{1}{2}\log(x^2+1) = \dfrac{1}{2}\log(y^2+1)+\log\sqrt{3}.<br />

    The question said in the begening express teh solution as an implicit equation relating x and y. That's the way it already is, right?
    Yes
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member Deadstar's Avatar
    Joined
    Oct 2007
    Posts
    722
    Quote Originally Posted by Hibachi View Post
    So \int\dfrac{x}{x^2 + 1}dx = \int\dfrac{y}{y^2+1} dy

    \therefore \;\;\; \dfrac{1}{2}\log(x^2+1) = \dfrac{1}{2}\log(y^2+1)+C

    when y = 0, x = 2

    \therefore \;\;\; \dfrac{1}{2}\log(3) = C

    \therefore \;\;\; \dfrac{1}{2}\log(x^2+1) = \dfrac{1}{2}\log(y^2+1)+\log\sqrt{3}.<br />

    The question said in the begening express teh solution as an implicit equation relating x and y. That's the way it already is, right?

    It's the right method yes but I think it should be \log(5) should it not? Since the x is squared.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Apr 2010
    Posts
    11
    Quote Originally Posted by Deadstar View Post
    It's the right method yes but I think it should be \log(5) should it not? Since the x is squared.
    You are right. Thanks, Deadstar.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Seperable Differential Equation
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: April 10th 2011, 06:25 PM
  2. seperable differential equation
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: October 3rd 2010, 05:21 PM
  3. Seperable Differential Equation
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: March 29th 2010, 07:14 PM
  4. Seperable Differential Equation
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: March 13th 2010, 06:41 AM
  5. Seperable differential equation
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: August 14th 2009, 07:22 AM

Search Tags


/mathhelpforum @mathhelpforum