What is the solution for xyy^'=x+yy^' ?

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

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xyy^'=x+yy^'

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d y}{d x} = (\frac{3 x ^{2}}{2 y})

1 - x^{2} y^{^{'}} = x 4 - y^{2}

f^{^{'}} (x) = f (x) + 4

\frac{d x}{d t} = k^{1} \cdot (1 - x) + k^{2} \cdot (1 - x)^{2}

y^{^{'}} = y \cdot y

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