x2 dy + (xy + y2) dx = 0
x2 dy = - (xy + y2) dx
⇒

dy

dx

= -

(xy+y2)

x2

... (1)
This is a homogeneous differential equation.
Such type of equations can be reduced to variable separable form by the substitution y
= vx.
Differentiating w.r.t. x we get,

d

dx

(y) =

d

dx

(vx) ⇒

dy

dx

= v + x

dv

dx

Substituting the value of y and

dy

dx

in equation (1), we get :
v + x

dv

dx

= -

[x×vx+(vx)2]

x2

⇒ x

dv

dx

= - v2 - 2v = - v (v + 2)
⇒

dv

v(v+2)

= -

dx

x

⇒

1

2

[

1

x

−

1

v+2

] dv = -

dx

x

Integrating both sides, we get:

1

2

[log v log(v + 2)] = - log x + logC
⇒

1

2

‌log(

v

v+2

) = log

C

x

⇒

v

v+2

= (

C

x

)2
Substituting v =

y

x

⇒

y x

y x +2

= (

C

x

)2
⇒

y

y+2x

=

C2

x2

⇒

x2y

y+2x

= D ... (2)
Now, it is given that y = 1 at x = 1.
⇒

1

1+2

= D ⇒ D =

1

3

Substituting D =

1

3

in equation (2), we get

x2y

y+2x

=

1

3

⇒ y + 2x = 3x2y
So, the required solution is y + 2x = 3x2y