Results 1 to 3 of 3
Like Tree1Thanks
  • 1 Post By Prove It

Math Help - variable separable!

  1. #1
    Member
    Joined
    Jul 2011
    Posts
    93

    variable separable!

     dy/dx=x^{-1}(2y^{\frac{1}{2}}-2y)

    have reduce to  \int\frac{1}{2y^{\frac{1}{2}}-2y}dy=\int\frac{1}{x}dx

    i have tried the u substitution but still unable to reduce the left-hand side to its simplest form, pls what is it am missing here?

    thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member kalyanram's Avatar
    Joined
    Jun 2008
    From
    Bangalore, India
    Posts
    142
    Thanks
    14

    Re: variable separable!

    Try y=u^2 substitution.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,964
    Thanks
    1008

    Re: variable separable!

    Quote Originally Posted by lawochekel View Post
     dy/dx=x^{-1}(2y^{\frac{1}{2}}-2y)

    have reduce to  \int\frac{1}{2y^{\frac{1}{2}}-2y}dy=\int\frac{1}{x}dx

    i have tried the u substitution but still unable to reduce the left-hand side to its simplest form, pls what is it am missing here?

    thanks!
    \displaystyle \begin{align*} \int{\frac{1}{2\sqrt{y} - 2y}\,dy} &= \int{\frac{1}{2\sqrt{y}\left( 1 - \sqrt{y} \right)}\,dy} \\ &= -\int{\frac{1}{1 - \sqrt{y}}\left( -\frac{1}{2\sqrt{y}}\right) dy} \end{align*}

    Now make the substitution \displaystyle \begin{align*} u = 1 - \sqrt{y} \implies du = -\frac{1}{2\sqrt{y}}\,dy \end{align*} and the integral becomes \displaystyle \begin{align*} -\int{\frac{1}{u}\,du} \end{align*}. I'm sure you can go from here
    Thanks from lawochekel
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Variable separable question.
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: August 19th 2011, 03:02 AM
  2. Replies: 4
    Last Post: June 25th 2010, 07:37 AM
  3. Replies: 2
    Last Post: May 3rd 2010, 12:38 AM
  4. separable extension and separable polynomial
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: August 30th 2009, 07:22 PM
  5. Replies: 5
    Last Post: January 10th 2009, 11:22 PM

Search Tags


/mathhelpforum @mathhelpforum