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First order differential equation

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  1. Offline

    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?
  2. Offline

    \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 ).
  3. Offline

    (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.
  4. Offline


    Wolfram says no constant/2.

    Not quite sure where that constant/2 has come from, I certainly don't get that either.
  5. Offline

    (Original post by starkrush)

    OP is correct, if I'm interpreting correctly.
  6. Offline

    (Original post by f1mad)

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

    And edited to reflect this clearer
  7. Offline

    (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.

  8. Offline

    That was to strikethrough the text!

    (Yes, this is very constructive maths help)


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