Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Methods that were used to solve the ODE

  1. Jan 26, 2005 #1
    Can anyone show me how to make this:

    [tex]xy=c\left( y+\sqrt{y^2-x^2}\right)[/tex]

    Look like this:

    [tex]y^2-cx=y\sqrt{ y^2-x^2}[/tex]

    These were given as the answers to an ODE problem.

    I'm assuming that they are equivalent because they are presented in the answer as:

    [itex]y^2-cx=y\sqrt{ y^2-x^2}[/itex], or equivalently [itex]xy=c\left( y+\sqrt{y^2-x^2}\right) [/itex]

    I got the second answer but I can't figure out how to algebraically manipulated it to make look like the first.

    Any takers?

    P.S. I'm not certain if it can even be done algebraically. It might have something to do with the methods that were used to solve the ODE. But if they are equivalent answers it should be possible to obtain one from the other using just algebra shouldn't it?

    My algebra skills suck! :yuck:
  2. jcsd
  3. Jan 26, 2005 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It's true iff it can be proven algebraically. (Hurrah for abstract mathematics)

    The method might become clear if you try solving both equations for c.

    Better hint in white:

    Try "rationalizing" the RHS -- multiply through by y - √(y2 - x2). (and handle specially the case where this actually equals zero)
  4. Jan 26, 2005 #3
    Ok, that was specifically the hint that I needed. :biggrin:

    I guess I need to start doing some algebra problems everyday like morning calisthenics or something. It's hard to do algebra if you aren't doing it every day. I pretty much know how to do all the tricks, I just can never figure out which trick to pull out of the bag to achieve my final goals!

    Thanks for the tip! :approve:
  5. Jan 26, 2005 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Just FYI, I solved for c first. I figured rationalizing was the way to go, because one expression had the radical in the numerator, and the other on the denominator.
  6. Jan 26, 2005 #5
    That little tid-bit of wisdom will definitely be useful to me in the future. To be perfectly honest, I'm so rusty with algebra that even that hint starting me in the face wouldn't have done much for me. But from now on when I see that situation I'll be thinking "rationalize!" :wink:

    I'm just so rusty it's embarrassing. :blushing:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Methods that were used to solve the ODE