Solve y=2px+p³y²

y = 2px + p³y², with p = dy/dx. It is solvable for x.

Solve for x

x = (y − p³y²)/(2p) = y/(2p) − p²y²/2

Differentiate w.r.t. y

dx/dy = 1/p:

1/p = 1/(2p) − (y/(2p²)) dp/dy − p²y − p y² dp/dy

1/(2p) + p²y = −(dp/dy)[y/(2p²) + py²]

Taking out common factors: (1 + 2p³y)/(2p) = −(dp/dy) · y(1 + 2p³y)/(2p²)

So (1 + 2p³y)[1/(2p) + (y/(2p²)) dp/dy] = 0.

General solution

From 1/p + (y/p²) dp/dy = 0: dp/p = −dy/y, so py = c, p = c/y.

Substituting in the given equation:

y = 2cx/y + c³y²/y³ = 2cx/y + c³/y → y² = 2cx + c³

y² = 2cx + c³

Singular solution

The factor 1 + 2p³y = 0 gives the envelope of this family. Differentiating y² = 2cx + c³ w.r.t. c: 0 = 2x + 3c², so c² = −2x/3. Then

y⁴ = c²(2x + c²)² = (−2x/3)(4x/3)² = −32x³/27

27y⁴ + 32x³ = 0 (singular solution)