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How to find verify solution to differential equation

  1. Sep 3, 2012 #1
    1. The problem statement, all variables and given/known data

    See attached photo

    2. Relevant equations



    3. The attempt at a solution
    I don't even know where to start on this problem. Usually, if I am given a possible solution to a DE, say y(t)=cos(t) as a solution to y''-y'+y=sin(t), I find whether it was a solution or not by manipulating it and then plugging it into the second equation that it may be a solution to and then seeing whether the resulting statement is true. For the ex. problem I give in this section, I would do this:
    y'= -sint
    y''= -cost
    then sub them in: -cost +sint +cost = sint
    and by simplifying get: sint = sint
    so for this case, it y=cost is a solution.
    But with the problem that is in the attached photo I have no idea what to do.

    Does anyone have an guidance as to how I can solve the attached problem?
    Thanks :)
     

    Attached Files:

  2. jcsd
  3. Sep 3, 2012 #2

    SammyS

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    Here's the "photo":

    attachment.php?attachmentid=50458&d=1346730413.jpg

    Well, find [itex]\displaystyle \frac{\partial}{\partial x} f(x-ct)[/itex] and then the partial with respect to x of that.

    Do the same with partials with respect to t.

    Then plug all that in.
     
  4. Sep 3, 2012 #3
    I don't have much experience with partial derivatives, am I on the right track here: the first partial derivative with respect to x, [tex]\displaystyle\frac{\partial}{\partial x} f(x-ct)[/tex] be: 1, therefore the second would be 0?
     
  5. Sep 3, 2012 #4
    Sorry for the formatting on that, I had not even heard of LateX before finding this site.
    Oh never mind this, it's fixed.
     
    Last edited: Sep 3, 2012
  6. Sep 3, 2012 #5

    SammyS

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    No.

    [itex]\displaystyle \frac{\partial}{\partial x} f(x-ct) = (1)\cdot f'(x-ct)\,,[/itex] because d/dx (x) = 1 .

    Similarly [itex]\displaystyle \frac{\partial}{\partial t} f(x-ct) = (-c)\cdot f'(x-ct)\,.[/itex]
     
  7. Sep 3, 2012 #6
    Oh, so that's what the reference to chain rule was about! Thank you!! :smile:
     
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