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

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

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y^'= x/(y-1)

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

y = 1 + x^{2} + c_{1} + 1, y = 1 - x^{2} + c_{1} + 1

Solution steps

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

Solve separable ODE:

y = 1 + x^{2} + c_{1} + 1, y = 1 - x^{2} + c_{1} + 1

y = 1 + x^{2} + c_{1} + 1, y = 1 - x^{2} + c_{1} + 1

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

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

\frac{d y}{d x} = y (2 x^{2} + 4)

(3 x^{2} y + 2 y) d x + (x^{3} + 2 x) d y = 0

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

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