Innocent looking , but not really

Question

Solve

y [ x³ y³ + 2x² -y ] d x + x³ [x y³ -2 ] d y = 0 , and y(1) =1

Answer 1

y [ x³ y³ + 2x² -y ] d x + x³ [x y³ -2 ] d y = 0 , and y(1) =1

SOLUTION

x³ y⁴ d x + 2 x² y d x − y² d x + x⁴ y³ d y −2 x³ d y = 0

Let us group by degree

x³ y⁴ d x + x⁴ y³ d y + 2 x² y d x −2 x³ d y = y² d x

x³ y³ ( y d x + x d y ) + 2 x² (y d x − x d y ) = y² d x

Let us divide by y²

x³ y ( y d x + x d y ) + 2 x² (y d x − x d y ) / y² = d x

Eventually dividing by x²

x y ( y d x + x d y ) + 2 (y d x − x d y ) / y² = d x / x²

(1/2) d (x y)² + 2 d ( x / y ) = d (- 1/ x )

(1/2) (x y)² + 2 ( x / y ) = (- 1/ x ) + ξ

and since y (1 ) = 1

(1/2) (1)² + 2 ( 1 ) = (- 1 ) + ξ

That is ξ = 7 / 2

And finally

x³ y³ + 4 x² + 2 y = 7x y