What is the y^'(x)=(9x^2)/(9y^2-7),y(1)=0 ?

The y^'(x)=(9x^2)/(9y^2-7),y(1)=0 is -(9x^2)/7

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y^'(x)=(9x^2)/(9y^2-7),y(1)=0

y^{^{'}} (x) = \frac{9 x ^{2}}{9 y ^{2} - 7}, y (1) = 0

- \frac{9 x ^{2}}{7}

Solution steps

\frac{9 x ^{2}}{9 y ^{2} - 7} = \frac{9 x ^{2}}{9 y ^{2} - 7}

Constant function results in a constant

y (1) = 0

= \frac{9 x ^{2}}{9 \cdot 0 ^{2} - 7}

Simplify

\frac{9 x ^{2}}{9 \cdot 0 ^{2} - 7} : - \frac{9 x ^{2}}{7}

= - \frac{9 x ^{2}}{7}

\int \frac{x + x ^{2}}{x ^{3}} d x

\int_{- 1}^{3} (x + 1)^{3} d x

(x^{4} + 2 y) d x - x d y = 0

0.2 x^{3} \cdot 3^{x}

x \to \infty lim (- 1^{2 x})

What is the y^'(x)=(9x^2)/(9y^2-7),y(1)=0 ?
The y^'(x)=(9x^2)/(9y^2-7),y(1)=0 is -(9x^2)/7