# why am I getting "x" in the denominator?

#### PG1995

Joined Apr 15, 2011
819
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

#### t_n_k

Joined Mar 6, 2009
5,455
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:
• PG1995