What is the solution for y^'=(y^2-x^2)/(xy) ?

The solution for y^'=(y^2-x^2)/(xy) is y=xsqrt(-2ln(x)+c_{1)},y=-xsqrt(-2ln(x)+c_{1)}

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y^'=(y^2-x^2)/(xy)

y^{^{'}} = \frac{y ^{2} - x ^{2}}{x y}

y = x - 2 ln (x) + c_{1}, y = - x - 2 ln (x) + c_{1}

Solution steps

y^{'} (x) = \frac{y ^{2} - x ^{2}}{x y}

Solve Bernoulli ODE:

y = x - 2 ln (x) + c_{1}, y = - x - 2 ln (x) + c_{1}

y = x - 2 ln (x) + c_{1}, y = - x - 2 ln (x) + c_{1}

3 x^{2} + \frac{8 x}{3} - 2

2 y^{^{'}} + y = 8 t^{2}

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

3 cos (x)

\frac{1}{2 - 3 x}, at x = 1

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