What is the (dy)/(dx)=(2x+y)/(3+3y^2-x),y(0)=0 ?

The (dy)/(dx)=(2x+y)/(3+3y^2-x),y(0)=0 is (2x)/(-x+3)

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(dy)/(dx)=(2x+y)/(3+3y^2-x),y(0)=0

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

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

Solution steps

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

Constant function results in a constant

y (0) = 0

= \frac{2 x + 0}{3 + 3 \cdot 0 ^{2} - x}

Simplify

\frac{2 x + 0}{3 + 3 \cdot 0 ^{2} - x} : \frac{2 x}{- x + 3}

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

\frac{d}{d x} (ln (sin (x)))

(3, 5), (- 2, - 20)

(1 - x)^{2}

y = 2 arcsin (3 x)

\int 24 x (6 - x^{2})^{\frac{1}{3}} d x

What is the (dy)/(dx)=(2x+y)/(3+3y^2-x),y(0)=0 ?
The (dy)/(dx)=(2x+y)/(3+3y^2-x),y(0)=0 is (2x)/(-x+3)