blackrose636
find d^2(y)/d(x^2) in terms of x and y (second derivative)
1-xy=x-y
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VeritasVosLiberabit
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\[1(\frac{ dy }{ dx })-(x)\frac{ dy }{ dx}+\frac{ dx }{dx }y=\frac{ dx }{ dx }-\frac{ dy }{ dx }\]
tell me how this looks it's been a while since I differentiated with this method.
VeritasVosLiberabit
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this would be the setup to solve for the first derivative
mahmit2012
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y=-1 & y"=0
blackrose636
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how did u get the second derivative?
@mahmit2012
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-( x*y' + y) = 1 -y'
Algebraic!
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solve for y'
Algebraic!
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differentiate the expression again
VeritasVosLiberabit
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I didn't do the second derivative yet. just do the first then repeat.
Algebraic!
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if you get another y' in your expression for y"
sub.s in your result from the first differentiation...
Algebraic!
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Does that make sense?
mahmit2012
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y=-1 y"=0 why you did not pay attention!
mahmit2012
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x#1
blackrose636
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i got y' is -y-1/x-1
then i tried doing quotient rule and thne i got confused
Algebraic!
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you fail at math @mahmit2012
Algebraic!
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k
Algebraic!
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you got y' right
mahmit2012
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every body sleeping !! cool
blackrose636
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yeah -y-1/x-1
Algebraic!
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now, f = ( -y-1) g = (x-1)
(f/g)' = f'g -fg'/g^2
f' = -y'
g' = 1
g^2 = (x-1)^2
Algebraic!
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plug them in and use the result from the first diff. : y' = -y-1/x-1
to sub.s in for y'
mahmit2012
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find y=-1 it's so easy !!!
Algebraic!
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got it @blackrose636 ?
blackrose636
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how did u even get y=-1?
Algebraic!
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gl @Callisto
mahmit2012
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y-xy=x-1
so y=-1 for all x#1
mahmit2012
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@Algebraic! did you get it !
Callisto
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Ah...
1 - xy = x-y
y-xy = x-1
y (1-x) = x-1
y = (x-1) / (1-x)
y=-1!
mahmit2012
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Thank you @Callisto
mahmit2012
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this is different seeing !