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

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

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

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

- \frac{2 x - 1}{3}

Solution steps

\frac{2 x - 1}{3 y ^{2} - 3} = \frac{2 x - 1}{3 y ^{2} - 3}

Constant function results in a constant

y (1) = 0

= \frac{2 x - 1}{3 \cdot 0 ^{2} - 3}

Simplify

\frac{2 x - 1}{3 \cdot 0 ^{2} - 3} : - \frac{2 x - 1}{3}

= - \frac{2 x - 1}{3}

\frac{d y}{d x} = \frac{2}{y}

\frac{d y}{d x} = \frac{( x ^{2} + 2 )}{y}

y^{^{'}} = e^{- y} sin (x), y (\frac{π}{2}) = \frac{1}{2}

y^{^{'}} + 2 y = 0, y (0) = 1.5

\frac{d y}{d x} + x^{2} y = 0

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