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

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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:

$\dfrac x y = \dfrac 1 y - y + C$

$\blacksquare$


Sources