What is the solution for y^'=(x-3)(y+1)^{2/3},y(0)=-1 ?

The solution for y^'=(x-3)(y+1)^{2/3},y(0)=-1 is y=(x^6}{216}-(x^5)/(12)+\frac{x^4)/2-x^3-1

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y^'=(x-3)(y+1)^{2/3},y(0)=-1

y^{^{'}} = (x - 3) (y + 1)^{\frac{2}{3}}, y (0) = - 1

y = \frac{x ^{6}}{216} - \frac{x ^{5}}{12} + \frac{x ^{4}}{2} - x^{3} - 1

Solution steps

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

Solve separable ODE:

y = \frac{x ^{6}}{216} - \frac{x ^{5}}{12} + \frac{x ^{4}}{2} - x^{3} - 1

y = \frac{x ^{6}}{216} - \frac{x ^{5}}{12} + \frac{x ^{4}}{2} - x^{3} - 1

\frac{d}{d x} (\frac{2 x + 3}{3 x ^{2} - 4})

(5 x - 4)^{4}, at x = 2

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What is the solution for y^'=(x-3)(y+1)^{2/3},y(0)=-1 ?
The solution for y^'=(x-3)(y+1)^{2/3},y(0)=-1 is y=(x^6}{216}-(x^5)/(12)+\frac{x^4)/2-x^3-1