Christos
  • Christos
calc, https://www.dropbox.com/s/o2v0t0mlnzr4djz/Screenshot%202015-12-07%2019.28.49.png?dl=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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hartnn
  • hartnn
you got dy/dx first?
Thadds2003
  • Thadds2003
so you would have to combine like terms......
hartnn
  • hartnn
i'll skip some steps, \(2x^2 + 2y^2 y' = 0 \\ x^2 + y^2 y' =0 ...(A)\\ y' = -x^2/y^2 ....(B) \\ From (A), 2x+ 2yy' + y^2 y'' = 0 \\ 2x+2y(-x^2/y^2) + y^2 y'' = 0 \) isolate y''

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Christos
  • Christos
can I just find dy/dx twice ?
Christos
  • Christos
dy/dx = -x^2/y^2
Christos
  • Christos
now (-x^2/y^2)' to find d^2y/dx^2
hartnn
  • hartnn
ofcourse. you need to use quotient rule then. I try to avoid that as much as possible, and use product rule wherever i can.
Christos
  • Christos
Ok because I dont understand the first method ..
hartnn
  • hartnn
go with quotient rule then :) we'll match our answers to verify!
Christos
  • Christos
ok!
hartnn
  • hartnn
let me know what u get, when u get...or ask if you get stuck... you'll have to use dy/dx = -x^2/y^2 again/
Christos
  • Christos
i got y/x
hartnn
  • hartnn
|dw:1449510476763:dw| got that first step? ^^
Christos
  • Christos
hmm isnt the quotient rule involving multiplication with the denominator
hartnn
  • hartnn
|dw:1449510653522:dw| ^^ thats a critical step, use of chain rule
hartnn
  • hartnn
now since we already know dy/dx |dw:1449510767793:dw|
Christos
  • Christos
isnt it like (( x^2)'*y^2 - x^2*(y^2)')/y^4
hartnn
  • hartnn
yes, my bad, i will change it
hartnn
  • hartnn
|dw:1449510913676:dw|
Christos
  • Christos
now if that equals = 0 we can get rid of the denominator
Christos
  • Christos
because the numerator as a whole is 0
hartnn
  • hartnn
|dw:1449510931322:dw| it does not equal 0 !
Christos
  • Christos
why not
hartnn
  • hartnn
|dw:1449510986144:dw|
hartnn
  • hartnn
because it equals d^2y /dx^2
Christos
  • Christos
ok
Christos
  • Christos
see thats where im confused
Christos
  • Christos
after that Idunno what i am doing
hartnn
  • hartnn
dy/dx = -x^2/y^2 again differentiating that w.r.t x will give left side as d^2y/dx^2 ....and not 0. once you get everything in x and y, just go for simplification!
Christos
  • Christos
how do i get rid of the dy/dx on the numerator
hartnn
  • hartnn
|dw:1449511184841:dw| we already know dy/dx in terms of x and y :)
Christos
  • Christos
aha! so you did it with substitution . In that case how is this method different from the first one?
Christos
  • Christos
Now that I got the gist of it im trying to understand your method as well
hartnn
  • hartnn
the only difference is that i avoided quotient rule :P
Christos
  • Christos
how ? ?
Christos
  • Christos
did you bring it up and set the power to - ?
hartnn
  • hartnn
i made sure not to keep anything in the denominator. x^2 + y^2 y' = 0 your next step was to isolate y' which made y^2 come in to denominator
hartnn
  • hartnn
my next step is to differentiate x^2+ y^2 y' =0 itself!
Christos
  • Christos
hmmmmmmmm
hartnn
  • hartnn
\(\color{#0cbb34}{\text{Originally Posted by}}\) @Christos did you bring it up and set the power to - ? \(\color{#0cbb34}{\text{End of Quote}}\) ^^ will also work
Christos
  • Christos
so 2x + 2yy' + y^2y'' = 0 => 2x + 2y(dy/dx) + y^2*(d^2y/dx^2) = 0
Christos
  • Christos
=>
Christos
  • Christos
2x + 2y(-x^2/y^2) + y^2*(d^2y/dx^2) = 0 and then I solve for d^2y/dx^2
Christos
  • Christos
is that how you did it ??
hartnn
  • hartnn
exactly! :)
Christos
  • Christos
Thanks a lot for the help ! :)
hartnn
  • hartnn
welcome ^_^

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