What is the y^'(x)=((1+3x^2))/(3y^2-6y),y(0)=1 ?

The y^'(x)=((1+3x^2))/(3y^2-6y),y(0)=1 is -(1+3x^2)/3

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y^'(x)=((1+3x^2))/(3y^2-6y),y(0)=1

y^{^{'}} (x) = \frac{( 1 + 3 x ^{2} )}{3 y ^{2} - 6 y}, y (0) = 1

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

Solution steps

Find

\frac{( 1 + 3 x ^{2} )}{3 y ^{2} - 6 y}

given

y (0) = 1

\frac{( 1 + 3 x ^{2} )}{3 y ^{2} - 6 y} = \frac{( 1 + 3 x ^{2} )}{3 y ^{2} - 6 y}

y^{2} : 1^{2}

y : 1

= \frac{1 + 3 x ^{2}}{3 \cdot 1 ^{2} - 6 \cdot 1}

Simplify

\frac{1 + 3 x ^{2}}{3 \cdot 1 ^{2} - 6 \cdot 1} : - \frac{1 + 3 x ^{2}}{3}

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

\frac{d}{d x} (a x cos (2 x) + b x sin (2 x))

\frac{\partial}{\partial x} (cos (x) \cdot sin (x))

\frac{d}{d x} (1 + e^{x^{2}})

\frac{d}{d x} (ln ∣ ln (x) ∣)

y = x^{(\frac{1}{l n ( x )})}

What is the y^'(x)=((1+3x^2))/(3y^2-6y),y(0)=1 ?
The y^'(x)=((1+3x^2))/(3y^2-6y),y(0)=1 is -(1+3x^2)/3