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

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

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

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

- \frac{5 + x}{4}

Solution steps

Find

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

given

y (0) = 2

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

y^{2} : 2^{2}

y : 2

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

Simplify

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

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

\frac{d x}{2 x} = d t

\frac{2 d y}{y} = \frac{2 cos ( 2 x )}{sin ( 2 x )} d x

\frac{d v}{d t} = 9.81 - 10 v, v (0) = 0

y^{3} \frac{d y}{d x} = 2 x^{4} y^{4} - 4 x^{4}

x^{3} \frac{d y}{d x} - sin (x) y = 0

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