- #1
Hawkingo
- 56
- 2
Homework Statement
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. Homework Equations [/B]
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)##
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.