Expert: Alon Mandes - 11/20/2008

Question
QUESTION: I have tried and i'm stuck, i dont know what kind of formula 2 use anymore..Cn u help me? I'm doing my revision nw..Thanks a million..A) If x^2-y^2=1, prove that y(d^2y/(dx)^2)+((dy)/(dx))^2=1, c)if y^2-2xy=2x, prove that (x-y)((d^2(y))/(dx^2)) + (2)((dy)/(dx))=((dy)/(dx))^2

ANSWER: Ok , let's review some of the facts regarding implicit function
derivatives :
1. d/dx {x} = 1
2. d/dy {y} = y'
3. d/dy {y'} = y'' = d^2y/dx^2
4. (fg)' = f'g+fg' -> d/dy {yy'} = y'^2+yy''

Now, to our exercise :

A) x^2-y^2=1
d/dx {x^2-y^2=1} = 2x-2yy'=0 . Deriving 2nd time gives :
d/dx {2x-2yy'=0} = 2-2y'^2-2yy''=0 -> 1-y'^2-yy''=0 -> 1=y'^2-yy''
-> 1=[dy/dx]^2-y[d^2y/dx^2]=1  Q.E.D

B) y^2-2xy=2x
d/dx {y^2-2xy=2x} = 2yy'-2y-2xy'=2 . Deriving 2nd time gives :
d/dx {2yy'-2y-2xy'=2} = 2y'^2+2yy''-2y'-2y'-2xy''=0 ->
y'^2+yy''-y'-y'-xy''=0 -> y'^2+y''(y-x)-2y'=0 -> y'^2=2y'-y''(y-x)
-> y'^2=2y'+y''(x-y) -> (x-y)[d^2y/dx^2]+2[dydx]=[dy/dx]^2
Q.E.D

Alon.

---------- FOLLOW-UP ----------

QUESTION: Thanks but I dont understand the part b where d/dx(2yy'-2y-2xy'=2), why ...-2y... Become ...-2y'-2y'...? Can you explain? Which formula do you use? Much appreciated

Answer
-2y didn't become -2y'-2y'. -2y becomes -2y'. The 2nd -2y' came from
the next form. Here, I will rewrite it more clearly :
d/dx {2yy'-2y-2xy'=2}
d/dx{2yy'}  -   d/dx{2y}   -   d/dx{2xy')          =     d/dx{2}
[2yy']'          - [2y]'      -  [2xy']'          =     0
[2y]'*y'+2y*[y']'  - 2y'    -  [2x]'*y'+2x*[y']'   =     0
2y'^2   +2yy''     - 2y'    -  2y'     +2xy''      =     0

2y'^2+2yy''-2y'-2y'-2xy''=0 -> y'^2=2y'+y''(x-y)

I hope this helps.

Alon.