What is the y^'(x)=(6x^2)/(6y^2-7),y(1)=0 ?

The y^'(x)=(6x^2)/(6y^2-7),y(1)=0 is -(6x^2)/7

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y^'(x)=(6x^2)/(6y^2-7),y(1)=0

y^{^{'}} (x) = \frac{6 x ^{2}}{6 y ^{2} - 7}, y (1) = 0

- \frac{6 x ^{2}}{7}

Solution steps

\frac{6 x ^{2}}{6 y ^{2} - 7} = \frac{6 x ^{2}}{6 y ^{2} - 7}

Constant function results in a constant

y (1) = 0

= \frac{6 x ^{2}}{6 \cdot 0 ^{2} - 7}

Simplify

\frac{6 x ^{2}}{6 \cdot 0 ^{2} - 7} : - \frac{6 x ^{2}}{7}

= - \frac{6 x ^{2}}{7}

e^{x} \frac{d y}{d x} + 3 e^{x} y = 2

y = x^{5^{x}}

\frac{\partial}{\partial y} (x sin (y) + z e^{x})

\int \frac{1}{64 e ^{- 5 x} + e ^{5 x}} d x

θ \to 0 lim (cot (π θ))

What is the y^'(x)=(6x^2)/(6y^2-7),y(1)=0 ?
The y^'(x)=(6x^2)/(6y^2-7),y(1)=0 is -(6x^2)/7