What is the solution for (dy)/(dx)=(3y^3)/(x^2),y(3)=1 ?

The solution for (dy)/(dx)=(3y^3)/(x^2),y(3)=1 is y=sqrt(-x/(-6+x))

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(dy)/(dx)=(3y^3)/(x^2),y(3)=1

\frac{d y}{d x} = \frac{3 y ^{3}}{x ^{2}}, y (3) = 1

y = - \frac{x}{- 6 + x}

Solution steps

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

Solve separable ODE:

y = - \frac{x}{- 6 + x}

y = - \frac{x}{- 6 + x}

(2 sin (x))^{^{'}}

(- 29, - 40), (- 28, 3)

x \to - \infty lim (x^{- 5})

\int (- 3 x^{- 4}) d x

\frac{\partial}{\partial x} (\frac{1}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}})

What is the solution for (dy)/(dx)=(3y^3)/(x^2),y(3)=1 ?
The solution for (dy)/(dx)=(3y^3)/(x^2),y(3)=1 is y=sqrt(-x/(-6+x))