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

The solution for 2xydy=(x^2-y^2)dx 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

2 x y d y = (x^{2} - y^{2}) d x

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

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 e^{a x} cos (b x) d x

y = 4 - x^{2}, (- 2, 2)

\frac{\partial}{\partial x} (sin (4 x^{2} y - 4 x y^{3}))

f (x) = \frac{6 x ^{2} + 8 x + 2}{x}

\int \frac{s ( s + 1 ) ^{2}}{s} d s

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