What is the solution for 2xyy^'-y^2=1 ?

The solution for 2xyy^'-y^2=1 is y=sqrt(c_{1)x-1},y=-sqrt(c_{1)x-1}

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2xyy^'-y^2=1

2 x y y^{^{'}} - y^{2} = 1

y = c_{1} x - 1, y = - c_{1} x - 1

Solution steps

2 x y y^{'} (x) - y^{2} = 1

Solve separable ODE:

y = c_{1} x - 1, y = - c_{1} x - 1

y = c_{1} x - 1, y = - c_{1} x - 1

y^{^{'}} = (x + 1) y^{2}, y (0) = \frac{1}{2}

u^{^{'}} = 5 e^{9 x - H (x)}

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

y^{^{'}} = y^{\frac{2}{3}}, y (0) = 0

y^{^{'}} = \frac{y}{cos ( x )}

What is the solution for 2xyy^'-y^2=1 ?
The solution for 2xyy^'-y^2=1 is y=sqrt(c_{1)x-1},y=-sqrt(c_{1)x-1}