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anonymous

  • 5 years ago

t^2dy/dt+2ty-y^5=0,t>0 solve the bernoulli equation by using a substitution v=y^1-n

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  1. anonymous
    • 5 years ago
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    can any one help me plz

  2. anonymous
    • 5 years ago
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    ok, first of all, you need your first fan

  3. anonymous
    • 5 years ago
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    above is the question

  4. anonymous
    • 5 years ago
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    hmm... let me think

  5. anonymous
    • 5 years ago
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    ok

  6. anonymous
    • 5 years ago
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    nah, sorry, TOO hard for me, cant figure it out, I am not good with those derivatives and stuff, sorry....

  7. anonymous
    • 5 years ago
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    ok

  8. anonymous
    • 5 years ago
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    you want to derive it?

  9. anonymous
    • 5 years ago
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    yes

  10. anonymous
    • 5 years ago
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    is the question:\[dy/dt (t^2 +2ty - y^5)\]?

  11. anonymous
    • 5 years ago
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    yes

  12. anonymous
    • 5 years ago
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    using v = y - n?

  13. anonymous
    • 5 years ago
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    i need to solve it by bernoulli equation substitution method

  14. anonymous
    • 5 years ago
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    v=y^1-n

  15. anonymous
    • 5 years ago
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    yes

  16. anonymous
    • 5 years ago
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    you can take y = v + n

  17. anonymous
    • 5 years ago
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    substitute Y in the equation and you'll get :\[t^2 + 2t(v+n) - (v+n)^5\]

  18. anonymous
    • 5 years ago
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    ok

  19. anonymous
    • 5 years ago
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    you want to derive with respect to t?

  20. anonymous
    • 5 years ago
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    yes

  21. anonymous
    • 5 years ago
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    then you'll treat (v+n) in the equation as constants

  22. anonymous
    • 5 years ago
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    and so, I think you'll get : = 2t + 2(v+n) -(v +n) = 2t +2v + 2n -v -n = 2t +v +n I guess :)

  23. anonymous
    • 5 years ago
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    since you're deriving with respect to t, then you'll ignore the others and treat them as constants. please correct me if I'm wrong

  24. myininaya
    • 5 years ago
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    y and v are functions of t though. Also, yummy the site andrius gave you it awesome and easy to read

  25. anonymous
    • 5 years ago
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    lol yeah, google it :)

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