What is the solution for y^'=(y-1)^2x^3 ?

The solution for y^'=(y-1)^2x^3 is y=-4/(x^4+4c_{1)}+(x^4)/(x^4+4c_{1)}+(4c_{1})/(x^4+4c_{1)}

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y^'=(y-1)^2x^3

y^{^{'}} = (y - 1)^{2} x^{3}

y = - \frac{4}{x ^{4} + 4 c _{1}} + \frac{x ^{4}}{x ^{4} + 4 c _{1}} + \frac{4 c _{1}}{x ^{4} + 4 c _{1}}

Solution steps

y^{'} (x) = (y - 1)^{2} x^{3}

Solve separable ODE:

y = - \frac{4}{x ^{4} + 4 c _{1}} + \frac{x ^{4}}{x ^{4} + 4 c _{1}} + \frac{4 c _{1}}{x ^{4} + 4 c _{1}}

y = - \frac{4}{x ^{4} + 4 c _{1}} + \frac{x ^{4}}{x ^{4} + 4 c _{1}} + \frac{4 c _{1}}{x ^{4} + 4 c _{1}}

\int \frac{2}{e ^{x}} d x

\int_{0}^{3} 1 + x^{4} + 2 x^{2} d x

(x + 4)^{^{'}}

\int \frac{e ^{x}}{1 + 1 + e ^{x}} d x

\frac{d}{d x} (e^{- x} cos (x) + e^{- x} sin (x))

What is the solution for y^'=(y-1)^2x^3 ?
The solution for y^'=(y-1)^2x^3 is y=-4/(x^4+4c_{1)}+(x^4)/(x^4+4c_{1)}+(4c_{1})/(x^4+4c_{1)}