Hello,
I don't agree with :
y' is correct.y''= 2x/ xln2x
y' is 1/u(x).
u'(x)=ln(2x)+1
Hence y''=-(ln(2x)+1)/(x²ln²(2x))
show that y=ln(ln2x) is a solution of y''+(y')^2 + 1/x(y')=0
This is what I have so far.
y= ln(ln2x)
y'= 1/(xln2x)
y''= 2x/ xln2x
0= 2x/ xln2x + (1/(xln2x))^2 + (1/x)(1/(xln2x))
0= 2x/ xln2x + 1/(xln2x)^2 + 1/ (x^2ln2x)
Then i assume i put the fractions together to get something like
0= (2x(xln2x)(x^2ln2x))/ (xln2x)^2(x^2ln2x))
I know there are more steps which ive tried but from there it gets
very messy.
Am i even on the right track or have i completely missed something?
Thanks.