Results 1 to 3 of 3

Math Help - First Order ODE

  1. #1
    Junior Member
    Joined
    Mar 2010
    Posts
    51

    First Order ODE

    This one gave me some trouble:

    2y(x^2-y+x)dx + (x^2-2y)dy=0

    I believe the integrating factor is e^(2x) but I may be wrong.

    Thanks on advance, sorry about lack of latex, I'll redo it when I get home and am not on my phone.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    I was bored so here is more than just the verification of the integrating factor.
    {\partial M\over\partial y}=2x^{2}-4y+2x

    {\partial N\over\partial x}=2x

    \frac{{M}_{y}-{N}_{X}}{N(x,y)}=\frac{2x^{2}-4y+2x-2x}{x^{2}-2y}=2

    {e}^{2\int dx}={e}^{2x}

    (2yx^{2}{e}^{2x}-2y^{2}{e}^{2x}+2yx{e}^{2x})dx+(x^{2}{e}^{2x}-2y{e}^{2x})dy

    f(x,y)=\int (x^{2}{e}^{2x}-2y{e}^{2x})dy=x^{2}{e}^{2x}y-y^{2}{e}^{2x}+g(x)

    x^{2}{e}^{2x}y-y^{2}{e}^{2x}+g(x)

    {\frac{d}{dx}}[x^{2}{e}^{2x}y-y^{2}{e}^{2x}+g(x)]=2yx^{2}{e}^{2x}-2y^{2}{e}^{2x}+2yx{e}^{2x}+g'(x)

    g'(x)=0

    g(x)=C

    x^{2}{e}^{2x}y-y^{2}{e}^{2x}+C
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2010
    Posts
    51
    That's exactly what I wanted, thanks a ton.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Re-writing higher order spatial derivatives as lower order system
    Posted in the Differential Equations Forum
    Replies: 11
    Last Post: July 27th 2010, 09:56 AM
  2. Replies: 1
    Last Post: October 27th 2009, 05:03 AM
  3. Replies: 2
    Last Post: February 23rd 2009, 06:54 AM
  4. Replies: 2
    Last Post: November 25th 2008, 10:29 PM
  5. Replies: 1
    Last Post: May 11th 2007, 04:01 AM

Search Tags


/mathhelpforum @mathhelpforum