What is the solution for y^'=(6y^3)/((3-2x)^2),y(0)=-1 ?

The solution for y^'=(6y^3)/((3-2x)^2),y(0)=-1 is y=-sqrt((2x)/(3(2x-1))-1/(2x-1))

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y^'=(6y^3)/((3-2x)^2),y(0)=-1

y^{^{'}} = \frac{6 y ^{3}}{( 3 - 2 x ) ^{2}}, y (0) = - 1

y = - \frac{2 x}{3 ( 2 x - 1 )} - \frac{1}{2 x - 1}

Solution steps

y^{'} (x) = \frac{6 y ^{3}}{( 3 - 2 x ) ^{2}}

Solve separable ODE:

y = - \frac{2 x}{3 ( 2 x - 1 )} - \frac{1}{2 x - 1}

y = - \frac{2 x}{3 ( 2 x - 1 )} - \frac{1}{2 x - 1}

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What is the solution for y^'=(6y^3)/((3-2x)^2),y(0)=-1 ?
The solution for y^'=(6y^3)/((3-2x)^2),y(0)=-1 is y=-sqrt((2x)/(3(2x-1))-1/(2x-1))