Correct answer: (4) y⁵ - y² – 1 = 0

Explanation:

(2x – 10y³) dy + ydx = 0

⇒ \(\frac{dx}{dy}\) + \((\frac{2}{y})\)x = 10y²

I.F. = \(e^{\int \frac{2}{y}dy}\) = \(e^{2 \ell n(y)}\) = y²

Solution of D.E. is

∴ x.y = ∫(10 y²)y².dy

xy² = \(\frac{10y^5}{5}\)+C 

⇒ xy² = 2y⁵ + C

It passes through (0, 1) → 0 = 2 + C ⇒ C = - 2

∴ Curve is xy² = 2y⁵ - 2

Now, it passes through (2, β)

2β² = 2β⁵ - 2 ⇒ β⁵ - β² - 1 = 0

∴ β is root of an equation y⁵ - y² – 1 = 0