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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

    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


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    Staff Emeritus
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    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.
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