Exact equation part 2

3.Exact Equations
A differential .equation :
M(x,y)dx+N(x,y)dy=0.
Is said to be exact if there exists a function f(x,y) such that
df(x,y)=M(x,y)dx+N(x,y)dy
Where M(x,y),N(x,y) and f(x,y) are continuous functions and have continuous first partial derivatives on some rectangle of the (x,y) plane .
Example: The equations ;
xdy+ydx=0
sin x cos y dy+cos x sin y dx=0
Are exact diff. eqs.
Since : d(xy)=xdy+ydx
d(sin x sin y)=sin x cos y dy+cos x sin y dx
Where as sin x cos y dy-cos x sin y dx=0 is not exact diff. eq .
Now if M(x,y)dx+N(x,y)dy=0. Is exact then there exists a function f(x,y) satisfy the condition
df(x,y)=?f/?x dx+?f/?y dy
To be ?f/?x=M and ?f/?y=N
i.e ?M/?y=?/?y (?f/?x)=(?^2 f)/?x?y=?/?x (?f/?y)=?N/?x ... (2)
this condition is sufficient for eq. to be exact .
Theorem : The diff. eq. M(x,y)dx+N(x,y)dy=0 is an exact differential equation iff ;
?M/?y=?N/?x

f(x,y) exists , so we get the solution by integrating
f(x,y)=???Mdx+g(y)=c?
Let
???M(x,y)dx=u?
Then:
?u/?x=M
?/?y (?u/?x)=?M/?y=?N/?x
By eq. (2) ,then:
?N/?x=?/?y (?u/?x)=?/?x(?u/?x)
N= ?u/?y+g (y)
By integration where g (y) is constant of integration is a function of y alone
Thus M(x,y)dx+N(x,y)dy=?u/?x dx+?u/?y dy+g^ (y)dy=du+dg=d(u+g)
Which is exact differential eq. .
Example: Solve the equation ; (x^2-4xy-2y^2 )dx+(y^2-4xy-2x^2 )dy =0
Here M(x,y)=x^2-4xy-2y^2 , then
?M/?y=-4x-4y
And ?N(x,y)=y?^2-4xy-2x^2 ,then
?N/?x=-4y-4x
Thus ; ?M/?y=?N/?x
And hence the eq. is exact .
f(x,y)=???Mdx+g(y)=???(x^2-4xy-2y^2 )dx+g(y) ??
=x^3/3-2x^2 y-2y^2 x+g(y)
?f/?y=-2x^2-4yx+dg/dy
So N(x,y)=-2x^2-4yx+dg/dy
y^2-4xy-2x^2=-2x^2-4yx+dg/dy
Then
y^2 = dg/dy
By integration
?dg=y?^2 dy
g(y)=y^3/3+c1
Then the solution is
x^3/3-2x^2 y-2y^2 x+y^3/3=c
Theorem : If the equation M(x,y)dx+N(x,y)dy=0 is not exact then there exists a function I(x,y) , Such that ;
I(x,y)[M(x,y)dx+N(x,y)dy]=0
Is exact and have the same solution , I(x,y) is called an integration Factor
Rule 1: Some an integrating factor can found by inspection