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

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

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

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

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

Solution steps

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

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 + \frac{3}{x}

3 sin (x), at x = \frac{π}{4}

y^{^{'}} + 3 x^{7} y^{9} + \frac{y}{x} = 0

y^{^{'''}} - 11 y^{^{''}} + 51 y^{^{'}} - 41 y = 0

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

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