Answer to Question #305032 in Differential Equations for She

Question

Find the complete solution of:

1.) (x ^ 2 y + y ^ 3 - y) d x + (x ^ 3 + xy ^ 2 + x) dy = 0

Answer 1

$$(x^2y+y^3-y)dx+(x^3+xy^2+x)dy=0\\ (x^2y+y^3-y)dx=-(x^3+xy^2+x)dy\\ (1/3+1)d[x^3y+x(y^3-y)]=-(1/3+1)d[xy^3+y(x^3+x)]\\ d[x^3y+x(y^3-y)]=-d[xy^3+y(x^3+x)]\\ \intop{d[x^3y+x(y^3-y)]}=\intop{-d[xy^3+y(x^3+x)]}\\ x^3y+x(y^3-y)=-xy^3+y(x^3+x)+C, C-const\\ x^3y+xy^3-xy=-xy^3+x^3y+xy+C\\ 2xy^3-2xy=C\\ xy^3-xy=C/2=C_1, C_1-const\\ x(y^3-y)=C_1\\ x=C_1/(y^3-y)$$

Answer: $x=C_1/(y^3-y), С_1-const$ and (x=0,y=0) is also a solution.