What is the y^'(x)=(15+3x^2)/(3y^2-12y),y(0)=2 ?

The y^'(x)=(15+3x^2)/(3y^2-12y),y(0)=2 is -(5+x^2)/4

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y^'(x)=(15+3x^2)/(3y^2-12y),y(0)=2

y^{^{'}} (x) = \frac{15 + 3 x ^{2}}{3 y ^{2} - 12 y}, y (0) = 2

- \frac{5 + x ^{2}}{4}

Solution steps

Find

\frac{15 + 3 x ^{2}}{3 y ^{2} - 12 y}

given

y (0) = 2

\frac{15 + 3 x ^{2}}{3 y ^{2} - 12 y} = \frac{15 + 3 x ^{2}}{3 y ^{2} - 12 y}

y^{2} : 2^{2}

y : 2

= \frac{15 + 3 x ^{2}}{3 \cdot 2 ^{2} - 12 \cdot 2}

Simplify

\frac{15 + 3 x ^{2}}{3 \cdot 2 ^{2} - 12 \cdot 2} : - \frac{5 + x ^{2}}{4}

= - \frac{5 + x ^{2}}{4}

y^{^{'}} - 2 t y = 0

sin (y) cos (x) d y = cos (y) sin (x) d x

\frac{d y}{d x} = \frac{1}{5} y cos^{2} (y)

\frac{d p}{d x} = k (a - p)

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

What is the y^'(x)=(15+3x^2)/(3y^2-12y),y(0)=2 ?
The y^'(x)=(15+3x^2)/(3y^2-12y),y(0)=2 is -(5+x^2)/4