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Basic Diff. Eq. problem

  1. Feb 2, 2015 #1
    1. The problem statement, all variables and given/known data
    Let xy''-y'=0. Try a solution of the form y=xr. Is this a solution for some r? If so, find all such r.

    2. Relevant equations
    xy''-y'=0

    y=xr

    3. The attempt at a solution
    The answer I came up with was: r*x-r*x=0, for all r.
    But in the solutions it says, "y = xr is a solution for r = 0 and r = 2."
    Is there some reason it would only work for r=0 and r=2?
     
  2. jcsd
  3. Feb 2, 2015 #2
    The answer you came up is not the correct one. Check the second derivative of y(x)=x^r.
     
  4. Feb 2, 2015 #3

    O_o

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    What you should do is plug in some concrete examples. I'll give you two and see if you can work out the rest: [tex]
    y = x^1 \\
    y' = 1 \\
    y'' = 0 \\

    \implies xy'' - y' = x(0) - 1 \neq 0[/tex]
    So you can see it doesn't work for r = 1. Now let's see why it does work for r = 2[tex]
    y = x^2 \\
    y' = 2x \\
    y'' = 2 \\

    \implies xy'' - y' = x(2) - 2x = 0[/tex]
    as required. Now try plugging in a few values of r less than zero and a few greater than 3 and see if you can see why they don't work.
     
  5. Feb 2, 2015 #4

    Thanks I see that now haha

    Thanks for your help; it makes sense now that I see it. Would there be away to find r directly or would I just have to try them all?
     
  6. Feb 2, 2015 #5

    O_o

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    You kind of have to think a bit outside the box to figure out which values of r are "different".

    If r = 0 then y is a constant so all if its derivatives are 0
    If r = 2 then its first derivative is a function of x but its second derivative is a constant
    if r < 0 or r >2 then all the derivatives are functions of x

    Try out the few odd ones then go from there. For instance, you don't need to try out r = 4 after you try out r = 3 because they follow the same pattern. Same with r = -1 and r = -2
     
  7. Feb 2, 2015 #6
    Ah yes I see. Alright thanks for your help!
     
  8. Feb 2, 2015 #7

    O_o

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    Hey there's actually a way you can do it without too much thinking: [tex]
    y = x^r \\
    y' = rx^{r-1} \\
    y'' = r(r-1)x^{r-2} \\

    \implies xy'' - y' = x[r(r-1)x^{r-2}] - rx^{r-1} = x^{r-1}r(r-1 -1) = x^{r-1}r(r-2)[/tex]

    Which equals zero for all x when r = 0 or r = 2. Sorry I didn't notice this before. A bit stressed out with my own midterms coming up. Cheers.
     
  9. Feb 3, 2015 #8

    epenguin

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    Homework Helper
    Gold Member

    It is not too difficult to get the general solution to this d.e. which is

    y = A x2 + B

    Which includes the solutions given.
     
  10. Feb 7, 2015 #9
    Wow thanks! I'll definitely keep this in mind when working on similar problems. No worries, good luck on your midterms!

    How did you get to that solution?
     
  11. Feb 7, 2015 #10

    epenguin

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    Homework Helper
    Gold Member

    First tell me if you've checked it works. Which if it does may even point you to how you could have solved it.
     
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