2yex∕y dx + (y - 2x ex∕y) dy = 0

dx

dy

=

2xe x y −y

2ye x y

... (1)
Let f (x , y) =

2xe x y −y

2ye x y

Then, (λx , λy) =

λ(2xe x y −y)

λ(2ye x y )

= λ0 [F (x,y)]
Thus, F(x, y) is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation.
Let x = vy
Differentiating w.r.t. y, we get

dx

dy

= v + y

dv

dy

Substituting the value of x and

dx

dy

in equation (1), we get
v + y

dv

dy

=

2vyev−y

2yev

=

2vev−1

2ev

or y

dv

dy

=

2vev−1

2ev

- v
or y

dv

dy

= -

1

2ev

or 2ev dv = -

dy

y

or ∫ 2ev . dv = - ∫

dy

y

or 2ev = - log |y| + C
Substituting the value of v, we get
2e

π

y

+ log |y| = C ... (2)
Substituting x = 0 and y = 1 in equation (2), we get
2e0 + log |1| = C ⇒ C = 2
Substituting the value of C in equation (2), we get
2e

x

y

+ log |y| = 2, which is the particular solution of the given differential equation.