Sol. The given differential equation is

y = px + p – p 3 ... (i)

Differentiating (i) w.r.t. x, we get

= p + x + –3p 2

⇒ p = p + x + –3p 2

⇒ (x + 1 –3p 2 ) = 0

⇒ = 0 ... (ii)

or, x + 1 – 3p 2 = 0 ... (iii)

Now, = 0 ⇒ p = C

Putting p = C in (i), we obtain y = Cx + C – C 3 which is the general solution of the given differential equation.

From (iii), we obtain

p = ±

Putting the value of p in (i), we get

y = ±

⇒ y = ± × (x + 1)

⇒ y

2 = (x + 1) 4 This is the singular solution of the given differential equation.