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}

\frac{d}{d x} (a^{x} - 4^{x})

x \to - 1 lim (2 x^{2} - 5 x - 1)

n = 0 \sum \infty \frac{4}{3 ^{n}}

\frac{d}{d x} (a x + b x + a y (x) + b y (x))

\int 3 sec^{6} (x) d 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