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Find y' if x[SUP]5y[/SUP]=y[SUP]2x[/SUP]

  1. Mar 14, 2013 #1
    1. The problem statement, all variables and given/known data

    Find y' if x5y=y2x.

    2. Relevant equations

    d(ln y)=dy/y.

    3. The attempt at a solution

    I know this involves the natural log, and I suspect I am supposed to take the ln of both sides in order to pop the exponents down in front, and then use d(ln y)=dy/y. But I can't really complete the thought. It's been a while!
  2. jcsd
  3. Mar 14, 2013 #2


    Staff: Mentor

    What do you get when you take the log (natural) of each side?
    IOW, what is ln(x5y)? What is ln(y2x)?
  4. Mar 14, 2013 #3
    I'm pretty sure you get (5y)(ln x)=(2x)(ln y), right?
  5. Mar 14, 2013 #4
    I know you can't isolate y, but I could atleast gather the y's on the left and the x's on the right, which gives me (5y)/(ln y)=(2x)/(ln x).
  6. Mar 14, 2013 #5
    Maybe instead I want (ln y)/(5y)=(ln x)/(2x), because then I have 5f(y)dy=2f(x)dx where f is the ln function.
  7. Mar 14, 2013 #6


    Staff: Mentor

    You don't need to isolate y. Just differentiate both sides of the first equation above implicitly.
  8. Mar 14, 2013 #7
    (5y)(dx/x) + (ln x)(5dy)=(2x)(dy/y) + (ln y)(2dx)
    (ln x)(5dy) - (2x)(dy/y)=(ln y)(2dx) - (5y)(dx/x)
    [5(ln x) - 2x/y]dy=[2(ln y) - 5y/x]dx
    dy/dx=[2(ln y) - 5y/x]/[5(ln x) - 2x/y]
  9. Mar 14, 2013 #8
    but i still get dy/dx in terms of y. Aren't I suppose to find it just in terms of x?
  10. Mar 14, 2013 #9


    Staff: Mentor

    This looks fine.

    All it asks is that you find y'. It doesn't say anything about it needing to be as a function of x alone.
  11. Mar 14, 2013 #10
    Hm. That can be done, though, right? Finding it in terms of x alone? Since I think that is what my professor wants, how would I do that?
  12. Mar 14, 2013 #11


    Staff: Mentor

    If that was what your professor wanted, why didn't he/she ask for that explicitly? There is nothing in the problem statement about solving for y' as a function of x alone.

    In any case, having the variables occur in both the base and the exponent makes it very difficult to separate them. The only thing I can think of that would be helpful is the Lambert W function.
  13. Mar 14, 2013 #12
    You're probably right, since this is a Calculus class and not anywhere near the level of a differential equations class, that's probably where we would stop.

    Although, I guess I could simplify to:

    dy/dx=[y[2x(ln y) - 5y]]/[x[5y(ln x) - 2x]]
  14. Mar 14, 2013 #13
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