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anonymous

  • 5 years ago

r^6 - 3r^4 + 3r^2 - 1 = 0 .. How to solve this differential equation ???

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  1. anonymous
    • 5 years ago
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    Factorise it if you want. Though it may be easier to let x = r^2 than factorise that (it's a standard expansion if you recognise it; if not you can find a factor (or three) by inspection). Then \[r = \pm \sqrt{x} \]

  2. anonymous
    • 5 years ago
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    ohh yea i get it! thanks alot !

  3. myininaya
    • 5 years ago
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    does this help r=1 multplicity 3 r=-1 multplicity 3

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  4. anonymous
    • 5 years ago
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    :) If you're interesting, it actually factorises it it's original for as: \[(r-1)^3(r+1)^3 \] Which I think is quite quaint.

  5. myininaya
    • 5 years ago
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    i tried to factor don't look at that first part where i factor r^4 of the first two terms

  6. myininaya
    • 5 years ago
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    i got lucky and synthetic division worked since all r all are rational zeros

  7. anonymous
    • 5 years ago
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    @myininaya. thanks it sure did help

  8. anonymous
    • 5 years ago
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    @newton. but u will have to factorise to get that answer, no ? we cant figure it out directly, can we ?

  9. anonymous
    • 5 years ago
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    \[(r-1)^3(r+1)^3 = 0 \iff r = \pm 1 \]

  10. anonymous
    • 5 years ago
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    yes yes i get it. what i meant to say was that we will have to factorise the original expression using division to reach to this answer ?

  11. anonymous
    • 5 years ago
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    Not necessarily using division, you could compare coefficients of something.

  12. anonymous
    • 5 years ago
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    (after spotting a few factors by inspection - in this case they are so obvious, but normally trying -2, -1, 1, 2 will get at least one)

  13. myininaya
    • 5 years ago
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    look i have something else if you didnt like synthetic

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  14. myininaya
    • 5 years ago
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    =(r^2-1)^3

  15. anonymous
    • 5 years ago
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    :D thanks alot. i get it

  16. myininaya
    • 5 years ago
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    thats alot of work for one problem lol

  17. anonymous
    • 5 years ago
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    haha yea that sure is. im highly obliged

  18. myininaya
    • 5 years ago
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    so none of the algebra I did needs explaining?

  19. anonymous
    • 5 years ago
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    nope its quite clear :) i completely get it, thanks to you :)

  20. myininaya
    • 5 years ago
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    :)

  21. anonymous
    • 5 years ago
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    @myininaya. could you please repost your cal0009.pdf attachment. it won't open for me. thanks

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