1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

DQ worksheet equation help

  1. Sep 22, 2013 #1
    1. The problem statement, all variables and given/known data
    This is from a worksheet...
    [itex]
    (\frac{y^2}{2}+ 2ye^x) + (y + e^x)\frac{_{dy}}{dx} = 0
    [/itex]

    2. Relevant equations



    3. The attempt at a solution

    After messing with this several times, i went through the through the text book for examples but nothing in the chapter matches with the exception of exact.. with a bit of manipulation i matched up the form. But it is not exact so that does not work. We have covered up separable, substitution, and exact and none seem to fit.

    [itex]

    \frac{\partial F}{\partial y}= y + 2e^x \neq \frac{\partial F}{\partial x} = e^x
    [/itex]
     
  2. jcsd
  3. Sep 22, 2013 #2

    rock.freak667

    User Avatar
    Homework Helper

  4. Sep 22, 2013 #3
    I see what you are saying but not how you can simplify the origional eq into what you got.

    I get

    [itex]
    y(\frac{y}{2}+2e^x) + (y+e^x)y' = 0
    [/itex]

    if you say that

    p(x) = [itex] (\frac{y}{2}+2e^x) [/itex]

    then p(x) is not going to integrate very nice.

    If i distribut the y'

    [itex]

    \frac{y^2}{2}+2ye^x + yy'+ y'e^x = 0

    [/itex]

    rearange a bit

    [itex]
    \frac{y^2}{2}+ yy' + 2ye^x + + y'e^x = 0
    [/itex]

    factor

    [itex]
    y(\frac{y}{2}+ y') + e^x(2y + y') = 0
    [/itex]

    but then what?
     
  5. Sep 22, 2013 #4
    I was thinking:

    [itex]

    (\frac{y^2}{2}+ 2ye^x)dx = -(y+e^x)dy

    [/itex]

    Then:

    [itex]

    \frac{xy^2}{2}+ 2ye^x = -\frac{1}{2}y^2 - ye^x + c
    [/itex]

    then to:
    [itex]
    \frac{xy^2}{2}+ 2ye^x + \frac{1}{2} y^2 + ye^x = c
    [/itex]
     
  6. Sep 23, 2013 #5
    You can't integrate like that since y is a function of x. You need to put it into standard form:

    [tex](y^2/2+2ye^x)dx+(y+e^x) dy=0=Mdx+Ndy[/tex]

    Then:

    [tex]\frac{1}{N}(M_y-N_x)=1=f(x)[/tex]

    The integrating factor is then [itex]e^{\int 1 dx}[/itex]

    Now multiply both sides of the standard-form by this integrating factor to make it exact then proceed to solve it using the technique of exact equations. Also, first check that it is indeed exact in case I made a mistake.
     
  7. Sep 23, 2013 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I did not know that Dixie Queen did this kind of calculation!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: DQ worksheet equation help
  1. Help with equation (Replies: 1)

Loading...