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

The solution for x(x+y)y^'+y(3x+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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x(x+y)y^'+y(3x+y)=0

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

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

Solution steps

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

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