What is the solution for y^'=(x^2-1)/(2x+xy) ?

The solution for y^'=(x^2-1)/(2x+xy) is y=-2+sqrt(x^2-2ln(x)+c_{1)+4},y=-sqrt(x^2-2ln(x)+c_{1)+4}-2

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y^'=(x^2-1)/(2x+xy)

y^{^{'}} = \frac{x ^{2} - 1}{2 x + x y}

y = - 2 + x^{2} - 2 ln (x) + c_{1} + 4, y = - x^{2} - 2 ln (x) + c_{1} + 4 - 2

Solution steps

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

Solve separable ODE:

y = - 2 + x^{2} - 2 ln (x) + c_{1} + 4, y = - x^{2} - 2 ln (x) + c_{1} + 4 - 2

y = - 2 + x^{2} - 2 ln (x) + c_{1} + 4, y = - x^{2} - 2 ln (x) + c_{1} + 4 - 2

y d x + (2 + x) d y = 0, y (- 1) = 4

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

y^{^{'}} + \frac{2 x}{1 + x ^{2}} y = \frac{- x}{1 + x ^{2}}

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

\frac{d y}{d x} = 10 x 25 - y^{2}

What is the solution for y^'=(x^2-1)/(2x+xy) ?
The solution for y^'=(x^2-1)/(2x+xy) is y=-2+sqrt(x^2-2ln(x)+c_{1)+4},y=-sqrt(x^2-2ln(x)+c_{1)+4}-2