Question 1199306
<pre>

{{{y*dx - x*dy}}}{{{""=""}}}{{{0}}}

That is easily solvable by separating the variables.  Its solution is y = Cx

But your professor wants you to tackle it as either an exact differential 
equation or one solvable by using an integrating factor to convert it to
an exact differential equation. 

First we check to see if the differential equation is exact:

We write it in the form {{{M*dx+N*dy}}}{{{""=""}}}{{{0}}}

{{{matrix(1,5,(y)*dx,""+"",(-x)*dy,""="",0)}}}

So M(x,y) = y,   N(x,y) = -x

We form the partial derivatives:

{{{M[y]=1}}}, {{{N[x]=-1}}}

Those are not equal, so the differential equation is not exact.

So we can try one of these integrating factors to multiply through by to
see if it comes out to be an exact differential equation:

{{{mu}}}{{{""=""}}}{{{matrix(4,1,"","","",e^( int(((M[y]-N[x])/N^"")*dx))))}}}{{{or}}}{{{mu}}}{{{""=""}}}{{{matrix(4,1,"","","",e^( int(((N[x]-M[y])/M^"")*dy))))}}}

Trying the first one:

{{{mu}}}{{{""=""}}}{{{matrix(4,1,"","","",e^( int(((1-(-1))/(-1)^"")*dx))))}}}{{{""=""}}}{{{matrix(2,1,"",e^int(-2,dx))}}}{{{""=""}}}{{{matrix(2,1,"",e^(-2int(dx)))}}}{{{""=""}}}{{{e^(-2x)}}}

So we multiply the original differential equation

{{{y*dx - x*dy}}}{{{""=""}}}{{{0}}}

 through by {{{mu=e^(-2x)}}}

{{{(e^(-2x)y)dx}}}{{{""+""}}}{{{(-e^(-2x)x)dy}}}{{{""=""}}}{{{0}}}

Let's see if that is exact.

Taking the partial of {{{(e^(-2x)y)dx}}} with respect to y givss {{{e^(-2x)}}}

Taking the partial of {{{(-e^(-2x)x)dy}}} with respect to x 
gives {{{-e^(-2x)(1)+x(2e^(-2x))}}}{{{""=""}}}{{{e^(-2x)(2x-1)}}}

They are not equal so this integrating factor did not work.  There is no
use to try the other one because the equation is symmetrical in x and y,
and it would be the same with the x's and y's swapped.

So we can't convert the given differential equation to an exact one by
the usual method.  That's because the method assumes it's possible to
find an integrating factor in terms of one variable only.  So it's not
possible.  So we can't do what your professor asked you to do.

So let's just solve it by separating the variables:

{{{y*dx - x*dy}}}{{{""=""}}}{{{0}}}

I like my dy to be at the first:

{{{-x*dy}}}{{{""=""}}}{{{-y*dx}}}

{{{x*dy}}}{{{""=""}}}{{{y*dx}}}

Divide both sides by {{{x*y}}}

{{{(x*dy)/(x*y)}}}{{{""=""}}}{{{(y*dx)/(x*y)}}}

{{{(cross(x)*dy)/(cross(x)*y)}}}{{{""=""}}}{{{(cross(y)*dx)/(x*cross(y))}}}

{{{dy/y}}}{{{""=""}}}{{{dx/x}}}

{{{int(dy/y)}}}{{{""=""}}}{{{int(dx/x)}}}

{{{ln(y)}}}{{{""=""}}}{{{ln(x)+ln(C)}}}

{{{ln(y)}}}{{{""=""}}}{{{ln(Cx)}}}

{{{y}}}{{{""=""}}}{{{Cx}}}

Edwin</pre>