Please help with this maths homework

Question

Solve the following differential equation:

(1-(2/y)+x)dy/dx + y=2/x - 1

Answer 1

APPROACH ONE ; SULUTION BY INSPECTION

[ y - 2/x +1 ] d x + [ 1 - 2/y + x ] d y = 0

y d x - 2 d x/ x + d x + d y - 2 d y/ y + x d y = 0

y d x + x d y - 2 d x/ x - 2 d y/ y + d x + d y = 0

d ( x y ) - d ( 2 ln x ) - d ( 2 ln y ) + d x + d y = 0

x y - 2 ln |x| - 2 ln |y| + x + y = ξ

x y - 2 ln | x y | + x + y = ξ

x y - ln ( x y)² + x + y = ξ

APPROACH TWO : EXACT

[ y - 2/x +1 ] d x + [ 1 - 2/y + x ] d y = 0 ( Eq. 1 )

Let M ( x, y ) = y - 2/x +1 and N ( x, y) = 1 - 2/y + x

Then ∂ M / ∂ x = ∂ N / ∂ y = 1 Therefore the given differential equation is an exact one.

This means that ( Eq. 1 ) is the total differential of Ω ( x , y ) = k

Then Ω ( x , y ) = ∫ [ y - 2/x +1 ] d x + f ( y )

Ω ( x , y ) = x y - 2 ln |x |+ x + f ( y )

Then ∂Ω / ∂ y = x + f ' ( y ) = 1 - 2/y + x ⇒ f ' ( y ) = 1 - 2/y ⇒ f ( y ) = y - 2 ln | y | Finally

the solution to the given differential equation is x y - 2 ln |x |+ x + y - 2 ln | y | = ξ

x y - ln ( x y)² + x + y = ξ

Answer 2

This is an EXACT ODE and can be put into std form of M(x,y)+N(x,y)dy/dx=0. Since M can be described as ψx and N as ψy we can find the integral of Mdx = ψ creating a constant of integration h(y) that can be solved for by finding ψy and knowing this result is N, h' can then be found and integrated directly to find h(y). The solution will then be Ψ = constant determined by initial conditions.