(1)
Let
F(x, y) = 2
2
x
y
x
y
xe
y
ye
−
Then
F(λx, λy) =
0
2
[F( , )]
2
x
y
x
y
xe
y
x y
ye
⎛
⎞
⎜
⎟
λ
−
⎜
⎟
⎝
⎠ =λ
⎛
⎞
⎜
⎟
λ⎜
⎟
⎝
⎠
Thus, F(x, y) is a homogeneous function of degree zero Therefore, the given
differential equation is a homogeneous differential equation To solve it, we make the substitution
x = vy (2)
Differentiating equation (2) with respect to y, we get
dx
dy  =
+
dv
v
y dy
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MATHEMATICS
404
Substituting the value of 
xand dx
dy  in equation (1), we get
dv
v
+y dy
 = 2
1
2
v
v ev
e
−
or
dv
y dy  = 2
1
2
v
v ev
v
e
− −
or
dv
y dy  =
2 ve1
−
or
2ev dv =
dy
y
−
or
2 ve
⋅dv
∫
 =
ydy
−∫
or
2 ev = – log |y| + C
and replacing v by 
x
y  , we get
2
yx
e  + log |y| = C