Where

y \neq = 0

and

x^{2} \neq = 1

, the separable differential equation gives

\frac{y d y}{2 y ^{2} + 1} = \frac{x d x}{x ^{2} - 1}, \frac{1}{4} lo g (2 y^{2} + 1) = \frac{1}{2} lo g ∣ x^{2} - 1∣ + C .

(1)

The requested point selects the implicit solution

2 y^{2} + 1 = (x^{2} - 1)^{2}

. The two real branches approaching that point are

y = \pm \frac{x}{2} x^{2} - 2, x > 2 .

(2)

There is an important initial value problem qualification: the differential equation as printed has no classical solution through the specified initial point. Its right side is undefined at

y = 0

, and these branches have an infinite one-sided derivative as

x ↓ 2

. They satisfy the differential equation on

x > 2

and extend continuously to the requested point. Multiplying by the vanishing denominator produces an implicit differential equation with a larger domain; it does not repair the original initial value problem.