What is the y^'(x)=(9x^2)/(9y^2-11),y(1)=0 ?

The y^'(x)=(9x^2)/(9y^2-11),y(1)=0 is -(9x^2)/(11)

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y^'(x)=(9x^2)/(9y^2-11),y(1)=0

y^{^{'}} (x) = \frac{9 x ^{2}}{9 y ^{2} - 11}, y (1) = 0

- \frac{9 x ^{2}}{11}

Solution steps

\frac{9 x ^{2}}{9 y ^{2} - 11} = \frac{9 x ^{2}}{9 y ^{2} - 11}

Constant function results in a constant

y (1) = 0

= \frac{9 x ^{2}}{9 \cdot 0 ^{2} - 11}

Simplify

\frac{9 x ^{2}}{9 \cdot 0 ^{2} - 11} : - \frac{9 x ^{2}}{11}

= - \frac{9 x ^{2}}{11}

\frac{d}{d h} (2 (2 + h)^{(2)} + 5 (2 + h) - 3)

y y^{^{'}} = \frac{t ^{2}}{( 1 + t ^{3} )}

\int x^{3} cosh (x) d x

\int \frac{3 x ^{3} - 8 x ^{2} + 10}{x ( x - 1 ) ^{3}} d x

(y^{2} - x y) d x - x^{2} d y = 0

What is the y^'(x)=(9x^2)/(9y^2-11),y(1)=0 ?
The y^'(x)=(9x^2)/(9y^2-11),y(1)=0 is -(9x^2)/(11)