Sol. We have,

x –2y –x = 0

⇒ =

⇒ =

This is a homogeneous differential equation. Putting y= vx and = v + x , it reduces to v + x = v ±

⇒ x = ±

⇒ = ±

⇒ log (v + ) = ±log x + log C

⇒ v + = Cx or, v + =

⇒ y + = Cx 2 or, y + = C

⇒ (Cx 2 –y) 2 = x 2 + y 2 or, (C –y) 2 = x 2 + y 2 ⇒ C

2 x 4 –2C x 2 y = x 2 or, C 2 –2cy = x 2 ⇒ C

2 x 2 –2Cy = 1 or, x 2 + 2Cy –C 2 = 0

⇒ x 2 – y – = 0 or, x 2 + 2Cy –C 2 = 0

Since the curve passes through (0, 2). Therefore,

x 2 – y – = 0 ⇒ – – = 0 ⇒ C = –

and, x 2 + 2Cy –C 2 = 0 ⇒ 4C –C 2 = 0 ⇒ C = 4

substituting these values of C in the respective equations, we obtain the equation of the required curve as

x 2 + 8y –16 = 0 ⇒ x 2 = –8 (y –2)

This represents a parabola opening downwards and cutting x- axis at (±4, 0). The shaded region in fig. represents the required area. We slice this region into vertical strips. For the approximating rectangle shown in fig. We have

Length = y, width= Δ x, Area = y Δ x

As the approximating rectangle can move between x = –4 and x = 4. So,

Required area =

=

= 2

= 2

= sq. units.