Results 1 to 4 of 4

Math Help - First Order Differential

  1. #1
    Member
    Joined
    Jan 2011
    Posts
    88

    First Order Differential

    2r(s^2+1)dr+(r^4+1)ds=0

    This is the second question in the textbook and I am able to do every question after it, so perhaps there is something simple that I am not seeing? The answer has no trig functions in it.

    (2r)/(r^4+1)dr+(s^2+1)ds=0

    The only way I can integrate the dr term is with trig functions, but according the the answers there is a simple way such that no trig functions are required.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member ILikeSerena's Avatar
    Joined
    Dec 2011
    Posts
    733
    Thanks
    121

    Re: First Order Differential

    Quote Originally Posted by quantoembryo View Post
    2r(s^2+1)dr+(r^4+1)ds=0

    This is the second question in the textbook and I am able to do every question after it, so perhaps there is something simple that I am not seeing? The answer has no trig functions in it.

    (2r)/(r^4+1)dr+(s^2+1)ds=0

    The only way I can integrate the dr term is with trig functions, but according the the answers there is a simple way such that no trig functions are required.
    Hi quantoembryo!

    After separation of variables, your equation should be:

    \frac{2r}{r^4+1}dr+\frac{1}{s^2+1}ds=0

    How would you integrate the ds term?
    Can you perhaps apply something similar to the dr term?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,829
    Thanks
    1602

    Re: First Order Differential

    The first term can be integrated with a substitution if you write \displaystyle \begin{align*} \frac{2r}{r^4 + 1} = \frac{2r}{\left( r^2 \right)^2 + 1} \end{align*} and let \displaystyle \begin{align*}  u = r^2 \implies du = 2r\,dr \end{align*}.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jan 2011
    Posts
    88

    Re: First Order Differential

    Ah. So simple. Thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: September 21st 2011, 04:37 PM
  2. Replies: 2
    Last Post: May 7th 2011, 11:26 AM
  3. Replies: 2
    Last Post: February 23rd 2009, 06:54 AM
  4. Replies: 2
    Last Post: November 25th 2008, 10:29 PM
  5. second order differential eq help
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 2nd 2008, 06:32 PM

Search Tags


/mathhelpforum @mathhelpforum