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Differential Equation 1

  1. Sep 24, 2010 #1

    [tex]y' = 6\frac{y\ln y}{x}[/tex]

    After separation and integration, I got

    [tex]\ln[\ln y] = 6\ln x + c_1[/tex]

    [tex]\Rightarrow \ln y = e^{\ln x^6 + c_1}[/tex]

    I am not sure how to get this into an explicit form for y, without it getting nasty. I know that there is usually a trick to make it look cleaner.

    Any thoughts?
  2. jcsd
  3. Sep 24, 2010 #2
    On a whim, I let [itex]c_1 = \ln c_2[/itex] and good things happen! I get,

    [tex]y = e^{\frac{x^6}{c_2}}[/tex]
  4. Sep 24, 2010 #3


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    Acutally, if you let [itex]c_1= ln(c_2)[/itex] then you would have [itex]ln(x^6)+ c_1= ln(x^6)+ ln(c_2)= ln(c_2x^6)[/itex] so that
    [tex]ln(y)= e^{ln(c_2x^6)}= c_2x^6[/tex]
    [tex]y= e^{c_2x^6}[/tex].

    That is, [itex]c_2[/itex] is multiplying [itex]x^6[/itex], not dividing it. But since it is simply an arbitrary constant, it really does not matter.
  5. Sep 24, 2010 #4
    Oops! It was a summation of logs! Not a difference ... Nice catch Halls! :smile:
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