What is the solution for 3xy+y^2+(x^2+xy)y^'=0 ?

The solution for 3xy+y^2+(x^2+xy)y^'=0 is y=(-x^2+sqrt(x^4+c_{1)})/x ,y=-(x^2+sqrt(x^4+c_{1)})/x

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3xy+y^2+(x^2+xy)y^'=0

3 x y + y^{2} + (x^{2} + x y) y^{^{'}} = 0

y = \frac{- x ^{2} + x ^{4} + c _{1}}{x}, y = - \frac{x ^{2} + x ^{4} + c _{1}}{x}

Solution steps

3 x y + y^{2} + (x^{2} + x y) y^{'} (x) = 0

Solve using substitution:

y = \frac{- x ^{2} + x ^{4} + c _{1}}{x}, y = - \frac{x ^{2} + x ^{4} + c _{1}}{x}

y = \frac{- x ^{2} + x ^{4} + c _{1}}{x}, y = - \frac{x ^{2} + x ^{4} + c _{1}}{x}

What is the solution for 3xy+y^2+(x^2+xy)y^'=0 ?
The solution for 3xy+y^2+(x^2+xy)y^'=0 is y=(-x^2+sqrt(x^4+c_{1)})/x ,y=-(x^2+sqrt(x^4+c_{1)})/x