anonymous
  • anonymous
t^2dy/dt+2ty-y^5=0,t>0 solve the bernoulli equation by using a substitution v=y^1-n
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
can any one help me plz
anonymous
  • anonymous
ok, first of all, you need your first fan
anonymous
  • anonymous
above is the question

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anonymous
  • anonymous
hmm... let me think
anonymous
  • anonymous
ok
anonymous
  • anonymous
nah, sorry, TOO hard for me, cant figure it out, I am not good with those derivatives and stuff, sorry....
anonymous
  • anonymous
ok
anonymous
  • anonymous
BTW, did you try google? http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/ode/first/bernoulli/bernoulli.html http://tutorial.math.lamar.edu/Classes/DE/Bernoulli.aspx http://www.sosmath.com/diffeq/first/bernouilli/bernouilli.html ???
anonymous
  • anonymous
you want to derive it?
anonymous
  • anonymous
yes
anonymous
  • anonymous
is the question:\[dy/dt (t^2 +2ty - y^5)\]?
anonymous
  • anonymous
yes
anonymous
  • anonymous
using v = y - n?
anonymous
  • anonymous
i need to solve it by bernoulli equation substitution method
anonymous
  • anonymous
v=y^1-n
anonymous
  • anonymous
yes
anonymous
  • anonymous
you can take y = v + n
anonymous
  • anonymous
substitute Y in the equation and you'll get :\[t^2 + 2t(v+n) - (v+n)^5\]
anonymous
  • anonymous
ok
anonymous
  • anonymous
you want to derive with respect to t?
anonymous
  • anonymous
yes
anonymous
  • anonymous
then you'll treat (v+n) in the equation as constants
anonymous
  • anonymous
and so, I think you'll get : = 2t + 2(v+n) -(v +n) = 2t +2v + 2n -v -n = 2t +v +n I guess :)
anonymous
  • anonymous
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
myininaya
  • myininaya
y and v are functions of t though. Also, yummy the site andrius gave you it awesome and easy to read
anonymous
  • anonymous
lol yeah, google it :)

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