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

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

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(dy)/(dx)=x(y-1)^2

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

y = - \frac{2}{x ^{2} + 2 c _{1}} + \frac{x ^{2}}{x ^{2} + 2 c _{1}} + \frac{2 c _{1}}{x ^{2} + 2 c _{1}}

Solution steps

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

Solve separable ODE:

y = - \frac{2}{x ^{2} + 2 c _{1}} + \frac{x ^{2}}{x ^{2} + 2 c _{1}} + \frac{2 c _{1}}{x ^{2} + 2 c _{1}}

y = - \frac{2}{x ^{2} + 2 c _{1}} + \frac{x ^{2}}{x ^{2} + 2 c _{1}} + \frac{2 c _{1}}{x ^{2} + 2 c _{1}}

y = (x^{3} - 16 x)^{10}, (4, 0)

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

\int_{0}^{5} e^{- 2 x} d x

f (x) = \frac{2}{x - 3}

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

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