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**1. The problem statement, all variables and given/known data**

The question is to solve the inexact equation by turning it into exact.the equation is ##( x + y + 4 ) d x + ( - x + y + 6 ) d y = 0##

Where "x" and "y" are variable.

2. Relevant equations

The question is to solve the inexact equation by turning it into exact.the equation is ##( x + y + 4 ) d x + ( - x + y + 6 ) d y = 0##

Where "x" and "y" are variable.

2. Relevant equations

1.(x+y+4)=m and (-x+y+6)=n

2.Integrating Factor =##\frac { 1 } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y }##

3.a=[(Integrating factor)*m]=$$\frac { ( x + y + 4 ) } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y }$$ and b=[(Integrating factor)*m]=$$\frac { ( x + y + 4 ) } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y }$$

4.##t = (x ^ { 2 } + y ^ { 2 } + 4 x + 6 y)##

**3. The attempt at a solution**

The given equation is a non exact homogeneous equation and is in the form

mdx+ndy = 0

now let's take

( x + y + 4 ) = m and ( - x + y + 6 ) = n

Here ##m x+n y##

=## x ^ { 2 } + x y + 4 x - x y + y ^ { 2 } + 6 y##

= ##x ^ { 2 } + y ^ { 2 } + 4 x + 6 y \neq 0##

So the integrating factor of the equation should be

I.F = ##\frac { 1 } { m x + n y } = \frac { 1 } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y }##

So now let's convert the inexact equation to exact by multiplying each term with integrating factor.so the new exact equation should be

##\frac { ( x + y + 4 ) } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y } + \frac { ( - x + y + 6 ) } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y } = 0##

=##a x + b y =0##

Where a=$$\frac { ( x + y + 4 ) } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y }$$ and b= $$\frac { ( x + y + 4 ) } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y }$$

Now the solution should be ##\int a d x + \int b d y = constant##

Where [in term "a" y is constant and the term "b" is free from x]

=##\int \frac { ( x + y + 4 ) d x } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y } = C##

(As there is no term in "b" free from x)

(##C##=constant)

So$$ \int \frac { ( x + y + 4 ) d x } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y } = $$

=$$\int \frac { x + 2 } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y } + \int \frac { ( y + 2 ) d x } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y }$$

[Now let's put ##(x ^ { 2 } + y ^ { 2 } + 4 x + 6 y) = t##

So ##( x + 2 ) d x = \frac { d t } { 2 }##]

=$$\frac { 1 } { 2 } \int \frac { d t } { t } + (\int \frac { ( y + 2 ) d x } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y } )$$

=$$\frac { 1 } { 2 } \log t + ( \int \frac { ( y + 2 ) d x } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y }) $$

=$$\log t ^ { 2 } + ( \int \frac { ( y + 2 ) d x } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y })$$

=$$\log ( x ^ { 2 } + ( y ^ { 2 } + 4 x + 6 y ) ^ { 2 } + (\int \frac { ( y + 2 ) d x } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y }) $$

So now ##\log ( x ^ { 2 } + y ^ { 2 } + 4 x + 6 y ) ^ { 2 } + ( \int \frac { ( y + 2 ) d x } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y }) ##= constant

Now I can't figure out what to do next.please help.