1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Suppose a, b, c are three real numbers such that

  1. Oct 18, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-10-17_22-28-10.png

    2. Relevant equations
    character equation

    3. The attempt at a solution
    Should I set a = ax2 b= bx c =c in the character equation?
     
  2. jcsd
  3. Oct 18, 2016 #2

    Ssnow

    User Avatar
    Gold Member

    Hi, you must start to substitute ##y_{1}(x)=x^{r_{1}}## in the differential equation and find ##r_{1}## that satisfy the equation ...,
    so start to calculate ##y'_{1}(x)## and ##y_{1}''(x)## ...

    The same for ##y_{2}(x)=x^{r_{2}}\ln{x}## ...
     
  4. Oct 18, 2016 #3
    y1 = xr1
    y1' = r1* xr1-1
    y1" = r1 (r1-1)* xr1-2 = ( r12 - r1) *xr1-2

    then plug in ODE?
     
  5. Oct 18, 2016 #4
    Then (I simply say r instead of r1)
    a(r2-r)xr +b r xr + cxr

    xr ( a(r2-r) + b r +c )
    And solve for r ?
     
  6. Oct 18, 2016 #5

    Mark44

    Staff: Mentor

    You can't solve the above, because they aren't equations!
    The equation you need to solve is a(r2-r)xr +b r xr + cxr = 0,
    which is equivalent to xr ( a(r2-r) + b r +c ) = 0
    Now solve for r.
     
  7. Oct 18, 2016 #6
    Oh yeah!! but I don't know a, b and c... there are four unknowns then
     
  8. Oct 18, 2016 #7

    Mark44

    Staff: Mentor

    But you have a relationship involving a, b, and c given in your problem statement.
     
  9. Oct 18, 2016 #8
    So I solve for r which is r = .... (in terms of a, b, c) probably be -b +/- sqrt (b^2 - 4 ac) / 2a

    Then use the relationship involving a, b, and c given in problem statement?

    But do I need one more equation?
     
  10. Oct 18, 2016 #9
    For r1, it is -B/2A which is -(b-a)/2a
    Do I end up wth the letter as coefficient, or any further?
     
  11. Oct 18, 2016 #10

    Mark44

    Staff: Mentor

    No it isn't.

    Start from this:
    xr ( a(r2-r) + b r +c ) = 0
    The part on the left has to be identically equal to zero; i.e., for all values of x.

    So either xr = 0 (can't be true)
    or a(r2-r) + b r +c = 0
    Solve this equation for r.
     
  12. Oct 18, 2016 #11

    Mark44

    Staff: Mentor

    Yes, this is what I get: r = -(b - a)/(2a), which is the same as (a - b)/(2a)
     
  13. Oct 18, 2016 #12
    Great, so this is it, for r1 ?? :)
     
  14. Oct 18, 2016 #13

    Mark44

    Staff: Mentor

    Yes.
     
  15. Oct 18, 2016 #14
    For r2,
    I have y2 = xr2 ln (x)
    y2' = xr2-1 +r2 xr2-1 ln (x)
    y2" = (r2-1)xr2-2 +r2 xr2-2 + (r2^2-r2)xr2-2 ln(x)

    Which then plug in to the ODE, i found r2 = -(b-a)/ 2a same as r1

    and whole ODE can be 0 if x =1
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted