What is the y^'(x)=(3x^2)/(3y^2-4),y(1)=0 ?

The y^'(x)=(3x^2)/(3y^2-4),y(1)=0 is -(3x^2)/4

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y^'(x)=(3x^2)/(3y^2-4),y(1)=0

y^{^{'}} (x) = \frac{3 x ^{2}}{3 y ^{2} - 4}, y (1) = 0

- \frac{3 x ^{2}}{4}

Solution steps

\frac{3 x ^{2}}{3 y ^{2} - 4} = \frac{3 x ^{2}}{3 y ^{2} - 4}

Constant function results in a constant

y (1) = 0

= \frac{3 x ^{2}}{3 \cdot 0 ^{2} - 4}

Simplify

\frac{3 x ^{2}}{3 \cdot 0 ^{2} - 4} : - \frac{3 x ^{2}}{4}

= - \frac{3 x ^{2}}{4}

What is the y^'(x)=(3x^2)/(3y^2-4),y(1)=0 ?
The y^'(x)=(3x^2)/(3y^2-4),y(1)=0 is -(3x^2)/4