A capricious Differential equation...

Question

Solve the differential equation.

y( x² + y² - 1 ) dx + x( x² + y² + 1 ) dy = 0

Answer 1

I came across this equation browsing the unanswered questions under Differential equations.

Was posted on 10/11/20 by Renato S, and it went unanswered for nine months..

Here is the solution in complete agreement with WA.

( yx² + y³ −y ) dx + ( x³ + x y² + x ) dy = 0

yx² dx + y³ dx −y dx + x³ dy + x y² dy+ x dy = 0

[ yx² dx + x³ dy ] + [ y³ dx+ x y² dy ] + x dy − y dx = 0

x² [ y dx + x dy ] + y² [ y dx + x dy ] + x dy − y dx = 0

[ y dx + x dy] [ x² + y² ] + x dy − y dx = 0

y dx + x dy + ( x dy − y dx ) / ( x² + y² ) = 0

d (x y ) + d ( tan^-1(y/x) ) = 0

x y + tan^-1(y/x) = λ, λ∈ℜ

Answer 2

Pretty sick! Wolfram Alpha gave solution as xy - tan^-1(x/y). The solution took about 18 steps with some of them requiring intermediate steps and clever tricks and views throughout. There must be a nice way to do this, but what I saw from WA, I can't unsee. Good Luck!