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Solutions to a Simple Differential Equation

  1. Aug 30, 2012 #1
    1. The problem statement, all variables and given/known data
    Determine the values of r for which the given differential equation has solutions of the form [itex]y=t^r[/itex] for [itex]t > 0[/itex].


    2. Relevant equations
    [itex]t^2 y'' - 13ty' + 48y = 0[/itex]


    3. The attempt at a solution
    The program (online) has a thing that walks me through the question. It first had me find the second and first derivatives of [itex]y=t^r[/itex], which are [itex](r-1)*(r)*(t^{r-2})[/itex] and [itex]rt^{r-1}[/itex], respectively. It then tells me to plug those into the original equation, which gives me [itex](t^2)(r-1)(r)(t^{r-2}) - (13)(t)(r)(t^{r-1}) + (48)(t^r)[/itex]. I can apparently simplify that to [itex]r^2 - 14r + 48 = 0[/itex], but I have no idea how that works. I have not continued with the problem.
     
    Last edited: Aug 30, 2012
  2. jcsd
  3. Aug 30, 2012 #2

    Mark44

    Staff: Mentor

    Continuing from where you left off:
    ##(t^2)(r-1)(r)(t^{r-2}) - (13)(t)(r)(t^{r-1}) + (48)(t^r) = 0##
    So ##(r^2 - r -13r + 48)t^r = 0##
    Or ##(r^2 - 14r + 48)t^r = 0##

    What must be happen for the equation above to be true?
     
  4. Aug 30, 2012 #3

    CAF123

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    Gold Member

    I get to a simplified form [itex] t^r( r^2 -14r +48) = 0. [/itex]
    I think you have mistyped 48 as 49?
     
  5. Aug 30, 2012 #4
    I'm trying to simplify it (that's what's getting me), and I can simplify it to [itex](t^r)(r-1) - 13 + 48[/itex]. I'm clearly doing something wrong, but I don't know what.
    It is 48, my apologies. Thanks!
     
  6. Aug 30, 2012 #5

    CAF123

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    Gold Member

    Know that, for example, [itex] t^2(r^2t^{r-2}) = r^2t^r. [/itex]
    Do you see whats happening here?
     
  7. Aug 30, 2012 #6
    Ah, that makes sense. Okay, using that, I now follow these steps to simplify:

    [itex](t^2)(r-1)(r)(t^{r-2}) - (13)(t)(r)(t^{r-1}) + (48)(t^r) = 0[/itex]
    [itex](t^r)(r-1)(r) - (13)(r) + (48)(t^r) = 0[/itex]
    [itex](t^{r}r^{2}-t^{r}r) - 13r + 48t^r = 0[/itex]
    [itex]t^{r}(r^{2}-r)-13r-48t^{r} = 0[/itex]
     
  8. Aug 30, 2012 #7

    CAF123

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    Close, [itex] -13trt^{r-1} = -13rt^r, [/itex] and now use [itex] t^r [/itex] as a common factor.
     
  9. Aug 30, 2012 #8
    Ah, okay, for some reason I canceled it out rather than writing [itex]t^r[/itex]. Copy-pasting my work from above and editing appropriately now gives me the below output, which should be correct.

    [itex](t^2)(r-1)(r)(t^{r-2}) - (13)(t)(r)(t^{r-1}) + (48)(t^r) = 0[/itex]
    [itex](t^r)(r-1)(r) - (13)(r)(t^r) + (48)(t^r) = 0[/itex]
    [itex](t^{r}r^{2}-t^{r}r) - 13rt^{r} + 48t^r = 0[/itex]
    [itex]t^{r}(-r^2 - r - 13r + 48) = 0[/itex]
    [itex]t^{r}(-r^2 - 14r + 48) = 0[/itex]

    With that, I need to solve for t, correct? Then I'll be able to easily find the values of r for which [itex]t > 0[/itex].
     
  10. Aug 30, 2012 #9

    Mark44

    Staff: Mentor

    What you show is incorrect, and you have lost sight of the fact that you're dealing with an equation. You are not simplifying the equation I showed, which was
    ## (r^2 - 14r + 48)t^r = 0##

    The goal is to solve for r, not t.

    This is not a hard problem. You've already done all of the hard work.

    Under what circumstances (i.e., for what r) does ## (r^2 - 14r + 48)t^r = 0##
     
  11. Aug 30, 2012 #10

    Mark44

    Staff: Mentor

    You have several errors in your work above, including a couple of sign errors.
     
  12. Aug 30, 2012 #11

    CAF123

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    Gold Member

    A stray minus cropped in the [itex] r^2 [/itex] term, but other than that you now have to solve for r. [itex] t^r ≠ 0 [/itex] (the graph is similar to an exponential). So the other term must equal 0.
     
  13. Aug 30, 2012 #12
    Okay, I apologize for being an idiot. I was looking at the [itex]t > 0[/itex] in the problem and thinking that's what I was solving for.

    The corrected final equation should be [itex]t^{r}(r^{2}-14r+48) = 0[/itex], as Mark44 said. Using that formula, r = 6 and r = 8. Submitted those and they are, in fact, correct.

    After reading through everything again, I apologize for not getting how to do it -- it should've been simple. Thank you all for your help, I really appreciate it!
     
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