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

Theorem
The first order ODE:
 * $(1): \quad x \rd y - y \rd x = \paren {1 + y^2} \rd y$

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

Proof
Rearranging, we have:
 * $\dfrac {y \rd x - x \rd y} {y^2} = -\paren {\dfrac 1 {y^2} + 1} \rd y$

From the Quotient Rule for Derivatives:
 * $\map \d {\dfrac x y} = \dfrac {y \rd x - x \rd y} {y^2}$

from which:
 * $\map \d {\dfrac x y} = -\paren {\dfrac 1 {y^2} + 1} \rd y$

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