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

  1. #1
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    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.
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  2. #2
    Moo
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    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)/(xln(2x))
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  3. #3
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    y''=-(ln(2x)+1)/(xln(2x))
    what does ln^2 equal:S is it the 2x squared or the ln?
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  4. #4
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    Quote Originally Posted by Cement View Post
    what does ln^2 equal:S is it the 2x squared or the ln?
    Moo means ln(2x) \cdot ln(2x)

    -Dan
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