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Homework Help: First-Order Linear ODE help?

  1. Sep 24, 2005 #1
    First-Order Linear ODE help???

    This is my first post here. I'm still getting used to LATEX syntax so please forgive any mistakes. My question is on a simple differential equation... it doesn't appear to be exact or homogeneous... where do I start?

    thanks in advance.

    DD86
     

    Attached Files:

  2. jcsd
  3. Sep 24, 2005 #2

    Tide

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    HINT: Look for an integrating factor.
     
  4. Sep 24, 2005 #3
    [tex]y'-4xy+2yln(y)/x=0[/tex]

    [tex]M(x,y)= -4xy \quad and \quad N(x,y)=2yln(y)/x[/tex]

    err.. I'm taking Calc 3 and diff eq at the same time so I hope my partials are right...

    [tex]M_{y}=-4x \quad and \quad N_{x}=0[/tex]

    IntegFactor is defined as [tex]I(x) = e^{\int (M_{y} - N_{x})/N dx} [/tex] according to my notes. yeah I'm still stuck. Could someone verify the partial derivatives though?

    EDIT: I now have [tex]y'+2y(-2x+ln(y)/x)=0[/tex] which looks nicer but im still at a loss for what to do next...

    What integrating factor am I looking for?
     
    Last edited: Sep 24, 2005
  5. Sep 24, 2005 #4

    saltydog

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    December, need to put it into the form:

    [tex]Mdx+Ndy=0[/tex]

    Thus for:

    [tex]y^{'}-4xy+\frac{2yln(y)}{x}=0[/tex]

    we'd have:

    [tex](2yln(y)-4x^2y)dx+xdy=0[/tex]

    But that's not happening for me either. Can't see a way to solve it.

    Tide . . . how about another hint? Somebody else too is ok.

    Edit: Oh yea, it's not linear because of the ln(y) term.
     
    Last edited: Sep 24, 2005
  6. Sep 24, 2005 #5
    DecemberDays86, This really doesnt have to do with anything, but, do you post on bodybuilding.com also?
     
  7. Sep 24, 2005 #6
    thanks for your input saltdog. My prof seems to like asking questions where you need to recognize "chain-ruled" things. For example, I have to get used to identifying a product of two terms as the derivative of some compossite function... not cool hehe
     
  8. Sep 25, 2005 #7

    Tide

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    How about this?

    Divide both sides of your equation by y, note that y'/y = (ln y)', multiply both sides by x and finally arrive at

    [tex]\frac {d}{dx} \left( x^2 \ln y \right) =4 x^3[/tex]

    then see where you can go from there!
     
  9. Sep 25, 2005 #8

    saltydog

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    Ok, I get it now. Thanks Tide. You too December . . . PF rocks. :smile:
     
  10. Sep 25, 2005 #9
    smart trick Tide!

    here's my mess of a solution:

    [tex]xy'+4x^2y-2ylny=0[/tex]
    [tex]y'-4xy=(-2ylny)/x[/tex]

    [tex]Let \quad y = e^{g(x)} \implies y'=g'e^{g(x)}[/tex]

    [tex]g'e^{g(x)}-4xe^{g(x)}= \frac {-2g(x)e^{g(x)}}{x}[/tex]

    divide out e^g(x) and rearrange

    [tex]g'+\frac {2}{x}g=4x, Let \quad P(x) = 2/x \quad and \quad Q(x)=4x[/tex]

    [tex]g(x) = e^{- \int P(x) dx}[\int Q(x)e^{\int P(x) dx} dx + C][/tex]

    [tex]\frac {1}{x^2}[x^4+C] \implies g(x)=x^2 + \frac {C}{x^2}[/tex]

    QUESTION: am I allowed to call the "c/x^2" term just another constant?

    back-sub [tex]y(x) = e^{g(x)} = e^{(x^2)}*e^{\frac {C}{x^2}}[/tex]


    Thanks for all of your help.
     
    Last edited: Sep 25, 2005
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