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

The solution for (dy)/(dx)=-(3-y)^2 is y= 1/(x+c_{1)}+(3x)/(x+c_{1)}+(3c_{1})/(x+c_{1)}

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(dy)/(dx)=-(3-y)^2

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

y = \frac{1}{x + c _{1}} + \frac{3 x}{x + c _{1}} + \frac{3 c _{1}}{x + c _{1}}

Solution steps

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

Solve separable ODE:

y = \frac{1}{x + c _{1}} + \frac{3 x}{x + c _{1}} + \frac{3 c _{1}}{x + c _{1}}

y = \frac{1}{x + c _{1}} + \frac{3 x}{x + c _{1}} + \frac{3 c _{1}}{x + c _{1}}

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

\frac{\partial}{\partial y} (y cos (x y))

x \to \infty lim (\frac{14}{x})

(\frac{( x - 6 )}{( x + 7 )})^{3}

x \to 0 lim (ln (\frac{x}{1 - x}))

What is the solution for (dy)/(dx)=-(3-y)^2 ?
The solution for (dy)/(dx)=-(3-y)^2 is y= 1/(x+c_{1)}+(3x)/(x+c_{1)}+(3c_{1})/(x+c_{1)}