why am I getting "x" in the denominator?

Discussion in 'Math' started by PG1995, Nov 14, 2011.

  1. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
    5
    Hi

    I was solving an example problem using a different method using a formula linked below. The answer I ended up has an "x" in the denominator which isn't there in solution the book presents. I've failed to figure out where I made a mistake. Could you please help me? Thanks a lot.

    Book's solution: http://img810.imageshack.us/img810/8342/examplek.jpg
    My Solution: http://img717.imageshack.us/img717/2970/img0003wu.jpg
    Formula used: http://img411.imageshack.us/img411/8596/exactequation.jpg

    With best regards
    PG
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    You went astray in some of the integration process and algebraic manipulations - you need to watch which integration variable is being applied at each step.

    M=e^{2y}-ycos(xy)
    N=2xe^{2y}-xcos(xy)+2y

    \int{M}\partial x=\int{(e^{2y}-ycos(xy))}\partial x=xe^{2y}-sin(xy)

    c=\int{M}\partial x+\int {\[ N-\frac{\partial}{\partial y} \int M \partial x \] }\partial y \\ =\int{M}\partial x+\int {\[ (2xe^{2y}-xcos(xy)+2y)}-\frac{\partial}{\partial y}(xe^{2y}-sin(xy)) \] \partial y \\ =\int{M}\partial x+\int {\[ (2xe^{2y}-xcos(xy)+2y)}-(2xe^{2y}-xcos(xy))\] \partial y \\ =\int{M}\partial x+ \int{2y} \partial y \\ = \int{M}\partial x+y^2 \\ =xe^{2y}-sin(xy)+y^2

    Hence

    xe^{2y}-sin(xy)+y^2=c

    or given c is a constant, we may write without loss of generality

    xe^{2y}-sin(xy)+y^2 + c=0

    to agree with the book solution
     
    Last edited: Nov 15, 2011
    PG1995 likes this.
  3. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
    5
    Thanks a lot, t_n_k. It was really helpful

    Best regards
    PG
     
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