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Differential Equation and Slope Field Questions.

  1. Apr 14, 2009 #1
    Hey,

    1. The problem statement, all variables and given/known data.
    I was reading through the Differential Equation portion of my textbook and didn't quite understand the following paragraph.
    The above paragraph seemed a little confusing since, conventionally,
    [tex]
    {y} = {f(x)}
    [/tex]

    Where,
    [tex]
    {{\frac {{d}^{}}{d{x}^{}}}{\Big[y\Big]}} = {{{f}^{\prime}}{(x)}}
    [/tex]

    so that it is understood that [itex]{{{f}^{\prime}}{(x)}}[/itex] like [itex]{f(x)}[/itex] is a function of only [itex]{x}[/itex]. However, in the above paragraph it is insisted that [itex]{{\frac {{d}^{}}{d{x}^{}}}{\Big[y\Big]}}[/itex] must be a function of [itex]{x}[/itex] and [itex]{y}[/itex]. In other words [itex]{{y}^{\prime}} = {{F}{(x, y)}}[/itex], why is that?

    2. Relevant equations.
    Knowledge Differential Equations and Slope Fields.

    3. The attempt at a solution.
    If we begin from the conventional notation that,
    [tex]
    {z} = {f(x, y)}
    [/tex]

    I run in to the problem that I don't know how to explicitly find [itex]{{z}^{\prime}}[/itex]. Specifically, I don't know how to differentiate [itex]{f(x, y)}[/itex] with respect to [itex]{x}[/itex] and [itex]{y}[/itex] simultaneously. So that the derivative is actually a function of [itex]{x}[/itex] and [itex]{y}[/itex], like [itex]{f(x, y)}[/itex]. How would I differentiate [itex]{z}[/itex] with respect to [itex]{x}[/itex] and [itex]{y}[/itex] simultaneously?

    I note however, that in this particular case we're talking about [itex]{{y}^{\prime}}[/itex] as opposed to [itex]{{z}^{\prime}}[/itex].
    Noting this I recall that we can rewrite,
    [tex]
    {\frac {dy}{dx}} = {{{f}^{\prime}}{(x)}}
    [/tex]

    as
    [tex]
    {dy} = {{{{f}^{\prime}}{(x)}}{dx}}
    [/tex]

    Where [itex]{y}[/itex] is found by integrating both sides of the above equation.

    However, when I try to do this with the equation given,
    [tex]
    {{y}^{\prime}} = {F(x, y)}
    [/tex]

    Which can be rewritten as,
    [tex]
    {dy} = {F(x, y)dx}
    [/tex]

    and when integrated is,
    [tex]
    {y} = {{\int_{}^{}}{F(x, y)dx}}
    [/tex]

    I find that I do not know how to evaluate the RHS. How would I evaluate it?

    Thanks,

    -PFStudent
     
  2. jcsd
  3. Apr 15, 2009 #2
    Hey,

    Still stuck on these questions, a little bit of help would be nice.

    Thanks,

    -PFStudent
     
  4. Apr 15, 2009 #3
    slope fields are the lamest things ever, use maple or something
     
  5. Apr 16, 2009 #4
    Hey,
    A more substantial and detailed reply would help, anyone?

    Thanks,


    -PFStudent
     
  6. Apr 16, 2009 #5
    Slope fields are a very important tool in analyzing, understanding and using many differential equations most notably non-linear ones: it's one thing to solve it, quite another to intutively understand what's going on. Why does my removal of one measly fish out of my grandpa's pond cause the entire fish population to collapse? You need to get Blanchard, Devaney, and Hall, "Differential Equations". Spend some time with it, six weeks, and you'll be tops telling us what to do. Mathematica has some nice tools for creating phase portraits (slope fields) and especially ver 7 has what, a really nice function I don't remember the name for generating very nice pictures of slope fields. Start with some simple ones: coupled set of two linear equations and generate the slope field, then go to more complicated ones.
     
  7. Apr 16, 2009 #6
    what text book are you using?
     
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