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Finding the degree of a deferential equation

  1. Sep 25, 2010 #1
    hi frinds im new to this forum pls guide me if im wrong at any place

    im very much confused abt finding the degree of deferential equation of various problems.I have got the defination of "degree "of deferential equation as "the highest exponent of the highest derivative "okk now im clear but actually working with problems im much wired can anyone help me pls
     
  2. jcsd
  3. Sep 25, 2010 #2
    Will try to help, what do you want to know?
     
  4. Sep 25, 2010 #3
    i wanna to know the degree of [d^2y/dx^2+(dy/dx)^3]^6/5=6y
     
  5. Sep 25, 2010 #4
    On the left hand side (LHS) of the equals sign, is that whole term divided by 5, or is that raised to the 6/5 power?
     
  6. Sep 25, 2010 #5
    no yaar the whole of the L.H.S IS raisaed to the power of 6/5 got it???
     
  7. Sep 25, 2010 #6
    Ok.... to the 6/5 power...

    I would raise both LHS and RHS to the fifth power.... this would cancel out the fractional power on the LHS. On the RHS we would now have (6y)^5.

    On the LHS, the differential terms in brackets will now be raised to the sixth power.
    If you now expand the LHS.... we will find that the highest degree of the highest derivative will be [(d2^y)/(dx^2)]^6.
     
  8. Sep 25, 2010 #7
    [(LHS term)^6/5]^5 = (LHS)^6 (The fractional power cancels out)
    Then expand LHS....
     
  9. Sep 25, 2010 #8
    How are you getting on with that Cindrilla? :)
    Any luck?
     
  10. Sep 25, 2010 #9
    hey i too had the same idea but in vain the answer to the above question is tht it's degree is 2
     
  11. Sep 25, 2010 #10
    sorry it's degree is 1 and order is 2
     
  12. Sep 25, 2010 #11
    the power of the total l.h.s has ben shifted to r.h.s such tht (6y)^5/6so we r left with only l.h.s as no exponent so dergree is 1 order is 2 but my question is y we should do this
     
  13. Sep 25, 2010 #12
    Ok... we have raised both LHS and RHS to the fifth power....
    On LHS we ought to now have this:
    [d^2y/dx^2+(dy/dx)^3]^6
    On RHS: (6y)^5

    Do you agree with that so far?
     
  14. Sep 25, 2010 #13
    ya i agree with wht my book & u r saying but y shouldn't the problem be solved in the way how im saying
     
  15. Sep 25, 2010 #14
    We don't have to worry about the power which dy/dx on the LHS, and y on the RHS, are raised to, we have to look at the power which d^2y/dx^2 is raised to, that is the highest order derivative... do you agree with that comment?
     
  16. Sep 25, 2010 #15
    Cindrilla, I would need to see your working out to be able to pinpoint where you may be going wrong...
    can give me some idea of how you are attempting to solve this yourself please?
     
  17. Sep 25, 2010 #16
    ya first i had taken out the power 5 on both sides so tht [L.H.S]^6=(6Y)^5 THEN naturally we have to expand however after expansion we will get tht the highest exponent of d^2y/dx^2 as 6 isn't it?then i would say tht the degree is 6 & order is 2
    then y is this method wrong??
     
  18. Sep 25, 2010 #17
    Cindrilla, that is correct.

    I read what you posted a little earlier:
    "the power of the total l.h.s has ben shifted to r.h.s such tht (6y)^5/6so we r left with only l.h.s as no exponent so dergree is 1 order is 2 but my question is y we should do this"

    The problem you pose invites us to get rid of the fractional 1/5 power only.
     
  19. Sep 25, 2010 #18
    I would say that raising both sides to the 5/6 power is mathematically correct, but then if we had to solve further for y, with the term on the RHS raised to the 5/6 power, it would be much harder to solve than just getting rid of the fractional power only...
     
  20. Sep 25, 2010 #19

    HallsofIvy

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    Science Advisor

    But take the 5/6 power of both sides of the equation changes the equation and changes the degree of the equation.

    If we have
    [tex]\left(\frac{d^2y}{dx^2}\right)^4= y[/itex]
    that is a fourth degree equation. To solve it, I would probably start by taking the fourth root of each side:
    [tex]\frac{d^2y}{dt^2}= \pm y^{1/4}[/itex]
    Those equations will have the same solutions but they are different equations. The fact that there is now no fourth power of the derivative does not change the fact that the orginal question is of degree 4.

    the problem given [itex][d^2y/dx^2+(dy/dx)^3]^6/5=6y[/itex], if multiplied out, would have the second derivative to the 6/5 power so, if you want to use the term "degree" here it would be 6/5.

    (I say "if you want to use the term "degree" here" because "degree" is normally used only for integer powers.)
     
  21. Sep 26, 2010 #20
    oh my frinds ur not getting my point at all okk now listen to my problem carefully
    when we had the given question lyk this stated below as



    { (d^2y/dx^2)+(dy/dx)^3}^6/5=6y
    then
    we would naturally take first 5th power on both sides so tht the 1/5th power of l.h.s will be cancelled to give (l.h.s)^6=(6y)^5
    okkk
    it is clear?
    now if we say tht the degree here is 6 then the answr is wrong but yy?
    wht is the mistake i wanna to know it



    ya i know tht we can take 5/6 th powr on both sides so tht l.h.s part will have no exponent whras r.h.s wil have (6y)^5/6 okkk
    this is correct as given in my book
     
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