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

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

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y^'=(1-x)/y

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d}{d x} (- 3 tan (3 x))

\frac{d}{d x} (- a sin (x))

d er i v a t i v e x + 1 + x^{2}

\int \frac{1}{x ^{3} - 512} d x

(2 y - \frac{1}{x} + cos (3 x)) \frac{d y}{d x} + \frac{y}{x ^{2}} - 4 x^{3} - 3 y sin (3 x) = 0

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