aaargh maths help!!

Discussion in 'Off-Topic Chat' started by horn-girlie, Mar 19, 2006.

  1. horn-girlie

    horn-girlie Member

    anyone know anything about 2nd order differential equations?!? yes, they really are as fun as they sound ... :(

    anyway, if there are any super duper maths whiz's out there then pletty please try this one ...

    Find a suitable intergrating factor to solve:
    x(dy/dx) + 3y = (e^x) /(x^2)

    think i must be using the wrong integrating factor thingy... :-? aaargh why would anyone do maths?!?!

    merci x
  2. flower girl

    flower girl Member

    Has tmp suddenly become the new way to cheat on maths home work? :tongue:
  3. horn-girlie

    horn-girlie Member

    have been trying to do it myself all weekend, honest ...
  4. drummergurl

    drummergurl Active Member

    it sounds like i should know what you're on about... having just done differential type stuff in AS maths...

    but i haven't a clue what your formula thingy is...

    sorry.. i am of no use whatsoever.
  5. chizzum

    chizzum Member

    havent a clue. Thats beyond me!
  6. i aint got a clue, but i av asked some mated but they said they did back in high school and cant remember how to do it. sorry
  7. The integrating factor
    First order differential equations of this form:.

    where P and Q are functions of x only, can be solved by use of an integrating factor, I where:

    By using the integrating factor, the differential equation can be changed into an exact differential equation. This is a differential equation where the main part of it is the exact derivative of a product, and the rest is easy to integrate.

    check out this website
  8. Steveo

    Steveo New Member

    thats a first order d.e. love.
  9. lottie4744

    lottie4744 Member

    don't know about the first bit but (e^x) /(x^2) would go too [(e^2x/2] divided by [x^3/3]
  10. wewizrobbed

    wewizrobbed Member


    I know I'm probably missing something but is that equ. not first order? Don't you need a y" ( not just y' ) for a 2nd order, where you then get the auxiliary equation and complementary function? :-?

    Also Lottie, is that right? Intergral of (f(x)/g(x)) isn't equal to int. f(x) / int. g(x).

    Sorry if this is all nonsense, only just up :)
  11. DaveR

    DaveR Active Member

    Maths used to be fun when it was just numbers. As soon as they started trying to make me count letters as well I lost the will to live. :(
  12. horn-girlie

    horn-girlie Member

    yep, thats true ;) oh well ... i guess it would help if i knew what topic i was on ... hehe
  13. The other son

    The other son Member

    In the words of hitler upon finding out that the russians where pretty close to taking over berlin..

  14. The other son

    The other son Member

  15. Daniel Sheard

    Daniel Sheard Member

    Your LHS works out as (1/x^2) d/dx{x^3 y}. You can solve it from there.

    The "a level" formulaic way is to get it in the form dy/dx + f(x) y =
    For yours, that would by dy/dx + 3/x y = (by dividing by x)
    Integrating factor is then R = e^{int 3/x dx}

    It's still only first order, though.
  16. Teflon1961

    Teflon1961 Member

    I'll guess at Db minor......:-?
  17. horn-girlie

    horn-girlie Member

    Thanks guys! have done it (i think :-? ) so thank you kindly to all you super clever lovely people :D xx