1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Y'=(y+9x)^2, (y+9x=v)

  1. May 30, 2008 #1
    This problem is out of Kreyszig's 9E of advanced engineering mathematics.

    I don't understand their closed form solution.
    Their solution ends up being
    y=3tan(3x+c)-9x.

    But when they were solving v=y+9x for y,
    then taking the derivative, they get:
    y'=v'-9, then set that =v^2

    --Thats the step I have issues with. They are taking the derivative,
    but with respect to what? If it is some dummy variable,
    it would be y'=v'-9x' (chain rule?). If it is y with respect to v,
    the derivative of y=v-9x would just be y'=v'. Unless x' with respect to
    a dummy variable is = 1, of course.

    Can someone please enlighten me?
     
  2. jcsd
  3. May 30, 2008 #2

    Astronuc

    User Avatar

    Staff: Mentor

    The problem simply takes dv/dx = d/dx(y+9x) = dy/dx+9 or y'+9 where ' denotes differentiation with respect to x in this case.
     
  4. May 30, 2008 #3
    ...oh man I was treating y as a constant and not as a variable. y is a function of x, so d/dx(y)=dy/dx...

    : smacks forehead : thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Y'=(y+9x)^2, (y+9x=v)
  1. Y'' = -y (Replies: 4)

  2. (1/2)y'' + y^2 = 1 (Replies: 9)

Loading...