2xy' + 2y = 2y y'
2xy' - 2y y' = -2y
y'(2x - 2y) = -2y
y' = y/(y-x)
y'' = ( (y-x)y' - y(y' - 1) )/(y-x)^2
= [ (y-x)(y/(y-x) ) - y( y/(y-x) - 1) ]/(y-x)^2
= [ y - y^2 /(y-x) + y ]/(y-x)^2
= [ y - y^2 /(y-x) + y ]/(y-x)^2 * (y-x)/y-x)
=[ 2y(y-x) - y^2 ]/(y-x)^3
= (y^2 - 2x)/(y-x)^3
you have to eliminate the y' in your y'' to get only x's and y's
check my algebra, I did not write it out on paper first.
implicitly differentiate f(x)
2xy= y ^2
find second derivative with relevance to x and y.
i get y''= -xy' - y/ (y-x)^2
this is wrong though. can someone please help me. thank you
3 answers
Thank you!!!! i got it :)
I didn't realize I had to omit the y' and I got y(-2x + y)/ (y-x)^3