Кто решит уравнение?

Жи есть ответ будет

Не я

arctg(y/x) = ln(√(x² + y²))

d/dx (arctg(y/x)) = d/dx (ln(√(x² + y²)))

(1 / (1 + (y/x)²)) ⋅ (y'x - y) / x² = (1 / √(x² + y²)) ⋅ (1 / (2√(x² + y²))) ⋅ (2x + 2yy')

(x² / (x² + y²)) ⋅ (y'x - y) / x² = (1 / (2(x² + y²))) ⋅ (2x + 2yy')

(y'x - y) / (x² + y²) = (x + yy') / (x² + y²)

y'x - y = x + yy'

y'x - yy' = x + y

y'(x - y) = x + y

y' = (x + y) / (x - y)