First Order ODE/x dy - y dx = (1 + y^2) dy

Theorem
The first order ODE:
 * $(1): \quad x \, \mathrm d y - y \, \mathrm d x = \left({1 + y^2}\right) \mathrm d y$

has the solution:
 * $\dfrac x y = \dfrac 1 y - y + C$

Proof
Rearranging, we have:
 * $\dfrac {y \, \mathrm d x - x \, \mathrm d y} {y^2} = - \left({\dfrac 1 {y^2} + 1}\right) \mathrm d y$

From the Quotient Rule for Derivatives:
 * $\mathrm d \left({\dfrac x y}\right) = \dfrac {y \, \mathrm d x - x \, \mathrm d y} {y^2}$

from which:
 * $\mathrm d \left({\dfrac x y}\right) = - \left({\dfrac 1 {y^2} + 1}\right) \mathrm d y$

Hence the result:
 * $\dfrac x y = \dfrac 1 y - y + C$