anonymous
  • anonymous
find d^2(y)/d(x^2) in terms of x and y (second derivative) 1-xy=x-y
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[1(\frac{ dy }{ dx })-(x)\frac{ dy }{ dx}+\frac{ dx }{dx }y=\frac{ dx }{ dx }-\frac{ dy }{ dx }\] tell me how this looks it's been a while since I differentiated with this method.
anonymous
  • anonymous
this would be the setup to solve for the first derivative
anonymous
  • anonymous
y=-1 & y"=0

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anonymous
  • anonymous
http://i.imgur.com/CMgMz.gif
anonymous
  • anonymous
mother of god...
anonymous
  • anonymous
how did u get the second derivative? @mahmit2012
anonymous
  • anonymous
-( x*y' + y) = 1 -y'
anonymous
  • anonymous
solve for y'
anonymous
  • anonymous
differentiate the expression again
anonymous
  • anonymous
I didn't do the second derivative yet. just do the first then repeat.
anonymous
  • anonymous
if you get another y' in your expression for y" sub.s in your result from the first differentiation...
anonymous
  • anonymous
Does that make sense?
anonymous
  • anonymous
y=-1 y"=0 why you did not pay attention!
anonymous
  • anonymous
x#1
anonymous
  • anonymous
i got y' is -y-1/x-1 then i tried doing quotient rule and thne i got confused
anonymous
  • anonymous
you fail at math @mahmit2012
anonymous
  • anonymous
k
anonymous
  • anonymous
you got y' right
anonymous
  • anonymous
every body sleeping !! cool
anonymous
  • anonymous
yeah -y-1/x-1
anonymous
  • anonymous
now, f = ( -y-1) g = (x-1) (f/g)' = f'g -fg'/g^2 f' = -y' g' = 1 g^2 = (x-1)^2
anonymous
  • anonymous
plug them in and use the result from the first diff. : y' = -y-1/x-1 to sub.s in for y'
anonymous
  • anonymous
find y=-1 it's so easy !!!
anonymous
  • anonymous
got it @blackrose636 ?
anonymous
  • anonymous
how did u even get y=-1?
anonymous
  • anonymous
anonymous
  • anonymous
y-xy=x-1 so y=-1 for all x#1
anonymous
  • anonymous
@Algebraic! did you get it !
Callisto
  • Callisto
Ah... 1 - xy = x-y y-xy = x-1 y (1-x) = x-1 y = (x-1) / (1-x) y=-1!
anonymous
  • anonymous
Thank you @Callisto
anonymous
  • anonymous
this is different seeing !

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