What is the solution for 2xydy+(x^2+y^2)dx=0 ?

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

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2xydy+(x^2+y^2)dx=0

2 x y d y + (x^{2} + y^{2}) d x = 0

y = - \frac{x ^{2}}{3} + \frac{c _{1}}{x}, y = - - \frac{x ^{2}}{3} + \frac{c _{1}}{x}

Solution steps

2 x y d y + (x^{2} + y^{2}) d x = 0

Solve Bernoulli ODE:

y = - \frac{x ^{2}}{3} + \frac{c _{1}}{x}, y = - - \frac{x ^{2}}{3} + \frac{c _{1}}{x}

y = - \frac{x ^{2}}{3} + \frac{c _{1}}{x}, y = - - \frac{x ^{2}}{3} + \frac{c _{1}}{x}

\int (\frac{x ( x - 2 )}{( x - 1 ) ^{3}}) d x

f (x) = sinh (4 x)

\frac{t d u}{d t} = t^{2} + 3 u

\frac{\partial}{\partial x} (- 6 x^{3} + 10 y^{2} - 4 x y)

\int (- 3 x + 4) e^{- x} d x

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