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}}

e^{x} (9 - 5 x + 5 x^{2})

\frac{d}{d x} (cos (- \frac{x}{3}))

t^{2} sin (3 t)

\int_{- 3}^{3} 12 - x^{2} - (x^{2} - 6) d x

x \to 2 lim (9 + (- 12))

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)}