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

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

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y^'=((x-1))/y

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d}{d x} (\frac{1}{3 ^{(- x^{2} - y^{2})} + 1})

\int 3 t^{- 4} (2 + 4 t^{- 3})^{- 7} d t

\int t^{3} t^{4} + 5 d t

x \to 0 lim (x^{2} + 3 x - 1)

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