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    I need to solve:

    Ln(y^x). dy/dx = 3x^2 . y
    Rearrange to get

    X.lny (dy) = 3x^2 (dx)
    Int (X.lny (dy)) = Int (3x^2 (dx))

    To get

    (lny)^2/2 = 3x^2/2 + C

    According to my mark scheme I should get C/2
    But I don't know why?
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    \int \ln y \not= \frac{(\ln y)^2}{2} - try integrating by parts (where u = ln y, and dv/dy = 1)

    Also, I don't see where your y from the first line has gone.
    And if you're finding c, then you'll have been given x and y values which you haven't stated in your post.

    Edit: And ignore the first two parts as they have been corrected below - next time try and keep the y in your working throughout, but equally (if not more so) it's my bad for not trying with it first ).
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    (Original post by claret_n_blue)
    ....
    To integrate ln(y)/y use substitution:

    let u= lny

    du/dy = 1/y

    du= dy/y

    -> I u du = u^2/2= (lny)^2/2.
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    http://www.wolframalpha.com/input/?i...dx+%3D+3x%5E2y

    Wolfram says no constant/2.

    Not quite sure where that constant/2 has come from, I certainly don't get that either.
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    (Original post by starkrush)
    ...
    Wrong.

    OP is correct, if I'm interpreting correctly.
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    (Original post by f1mad)
    Wrong.

    OP is correct, if I'm interpreting correctly.
    Ninja edited 1 minute before you replied

    And edited to reflect this clearer
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    (Original post by starkrush)
    Ninja edited 1 minute before you replied

    And edited to reflect this clearer
    Last edited by f1mad; 13 Minutes Ago at 16:46

    Last edited by starkrush; 1 Minute Ago at 16:57.

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    That was to strikethrough the text!

    (Yes, this is very constructive maths help)
 
 
 
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