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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

(x + 5) e^{2 x} + 7

\frac{d y}{d t} = 5 \cdot (y - t^{2})

\frac{\partial}{\partial x} (\frac{1}{( 1 - x ) ^{2}})

sin (\frac{t}{2})

\frac{d}{d x} (cos (\frac{x + 1}{x - 1}))

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