Answer to Question #305049 in Differential Equations for One

Differential Equations

Question #305049

Find the complete solution of xdy + (x ^ 3 - xy ^ 2 - y) dy = 0

Expert's answer

xdy+(x^3-xy^2-y)dy=0\\ (x^3-xy^2-y)dy=-xdy\\ (1-1/2-1/3)d[x^3y+xy^3+y^2]=-d[xy]\\ \intop{1/6d[x^3y+xy^3+y^2]}=\intop{-d[xy]}\\ \intop{d[x^3y+xy^3+y^2]}=\intop{-6d[]}\\ x^3y+xy^3+y^2=-6xy+C, C-const\\ x^3y+xy^3+y^2+6xy=C, C-const\\

Answer: $x^3y+xy^3+y^2+6xy=C, C-const$ and (x=0,y=0) is also a solution.

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