(x2–y2)dx + 2xydy = 0
⇒

dy

dx

=

y2−x2

2xy

... (1)
It is a homogeneous differential equation.
Let y = vx ...(2)
∴

dy

dx

= v + x

dv

dx

... (3)
Substituting (2) and (3) in (1), we get:
v + x

dv

dx

=

v2x2−x2

2xvx

v + x

dv

dx

=

x2v2−1

2vx2

-

v2−1

2v

2v2+2vx

dv

dx

= v2 - 1
2vx

dv

dx

= - v2 - 1
(

2v

v2+1

) dv = -

dx

x

Integrating both sides, we get:
∫

2v

v2+1

dv = - ∫ (

1

x

) dx
log |v2+1| = - log |x| + log C
log |v2+1| = log |

C

x

|
v2 + 1 =

C

x

xv2 + 1 = C
x |(

y

x

)2+1| = C
y2+x2 = Cx ... 4
It is given that when x = 1, y = 1
(1)2+(1)2 = C(1)
C = 2
Thus, the required solution is y2+x2 = 2x.
OR
We need to solve the following differential equation

dy

dx

=

x(2y−x)

x(2y+x)

dy

dx

=

2y−x

2y+x

... (1)
It is a homogeneous differential equation.
Let y = vx ...(2)
∴

dy

dx

= v + x

dv

dx

... 3
Substituting (2) and (3) in (1), we get:
v + x

dv

dx

=

x(2v−1)

x(2v+1)

x

dv

dx

=

v−1

2v+1

- v
x

dv

dx

=

−2v2+v−1

2v+1

(

2v+1

−2v2+v−1

) dv = (

1

x

) dx
(

2v+1

2v2−v+1

) dv = (−

1

x

) dx
Integrating both sides,
∫

1

2

(

4v−1+3

2v2−v+1

) dv = ∫ (−

1

x

) dx
∫

1

2

(

4v−1

2v2−v+1

) dv + ∫

3

2

(

1

2v2−v+1

) dv = ∫ (−

1

x

) dx
∫

1

2

(

4v−1

2v2−v+1

) dv + ∫

3

4

|

1

v2− v 2 + 1 2

| dv = f (−

1

x

) dx
1/2 log |2v2−v+1| +

3

4

‌∫(

1

v2− v 2 + 1 16 + 7 16

) dv = - log |x| + C
1/2 log |2v2−v+1| +

3

4

∫

dv

(v− 1 4 )2+( √7 4 )2

= - log |x| + C
1/2 log |2v2−v+1| +

3

4

×

4

√7

tan−1(

v− 1 4

√7 4

) = - log |x| + C
1/2 log |2v2−v+1| +

3

√7

tan−1(

4v−1

√7

) = C - log |x|
Put v =

y

x

1

2

log |2(

y

x

)2−(

y

x

)+1| +

3

√7

tan−1(

4y x −1

√7

) = C - log |x|

1

2

log |

2y2−xy+x2

x2

| +

3

√7

tan−1(

4y−x

√7x

) = C - log |x| ... 4
Now y = 1 when x = 1

1

2

log |

(2)(1)2−1(1)+12

12

| +

3

√7

tan−1|

(4)(1)−1

√7(1)

| = C - log |1|

1

2

log 2 +

3

√7

tan−1(

3

√7

) = C ... (5)
Therefore, form (4) and (5) we get:

1

2

log |

2y2−xy+x2

x2

| +

3

√7

tan−1(

4y−x

√7x

) =

1

2

log 2 +

3

√7

tan−1(

3

√7

) - log |x|

1

2

log |

2y2−xy+x2

x2

| -

1

2

log 2 + log |x| =

3

√7

[tan−1

3

√7

−tan−1(

4y−x

√7x

)]

1

2

log |

2y2−xy+x2

2x2

.x2| =

3

√7

tan−1(

3x−4y+x √7x

1+ 3(4y−x) 7x

)

1

2

log |

2y2−xy+x2

2

| =

3

√7

tan−1(

4(x−y) √7x

7x+12y−3x 7x

)

1

2

log |

2y2−xy+x2

2

| =

3

√7

tan−1(

√7(x−y)

x+3y

)