# Math Help - show that y=ln(ln2x) is a solution of y''+(y')^2 + 1/x(y')=0

1. ## show that y=ln(ln2x) is a solution of y''+(y')^2 + 1/x(y')=0

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.

2. Hello,

I don't agree with :

y''= 2x/ xln2x
y' is correct.

y' is 1/u(x).

u'(x)=ln(2x)+1

Hence y''=-(ln(2x)+1)/(x²ln²(2x))

3. y''=-(ln(2x)+1)/(x²ln²(2x))
what does ln^2 equal:S is it the 2x squared or the ln?

4. Originally Posted by Cement
what does ln^2 equal:S is it the 2x squared or the ln?
Moo means $ln(2x) \cdot ln(2x)$

-Dan