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

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

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(x+1)^{2/3}(dy)/(dx)=y^{1/3}

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

y = \frac{2 2 27 x + 27 + 27 c _{1} ( x + 1 ) ^{\frac{2}{3}} + 9 c _{1}^{2} ( x + 1 ) ^{\frac{1}{3}} + c _{1}^{3}}{3 3}, y = - \frac{2 2 27 x + 27 + 27 c _{1} ( x + 1 ) ^{\frac{2}{3}} + 9 c _{1}^{2} ( x + 1 ) ^{\frac{1}{3}} + c _{1}^{3}}{3 3}

Solution steps

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

Solve separable ODE:

y = \frac{2 2 27 x + 27 + 27 c _{1} ( x + 1 ) ^{\frac{2}{3}} + 9 c _{1}^{2} ( x + 1 ) ^{\frac{1}{3}} + c _{1}^{3}}{3 3}, y = - \frac{2 2 27 x + 27 + 27 c _{1} ( x + 1 ) ^{\frac{2}{3}} + 9 c _{1}^{2} ( x + 1 ) ^{\frac{1}{3}} + c _{1}^{3}}{3 3}

y = \frac{2 2 27 x + 27 + 27 c _{1} ( x + 1 ) ^{\frac{2}{3}} + 9 c _{1}^{2} ( x + 1 ) ^{\frac{1}{3}} + c _{1}^{3}}{3 3}, y = - \frac{2 2 27 x + 27 + 27 c _{1} ( x + 1 ) ^{\frac{2}{3}} + 9 c _{1}^{2} ( x + 1 ) ^{\frac{1}{3}} + c _{1}^{3}}{3 3}

f (x) = \frac{x}{x ^{2} + 49}

\int_{0}^{1} \frac{4}{1 + t ^{2}} d t

ln (1 + x^{2})

x^{2} + 7, 2 x^{2} + 2 x - 1

\int \frac{1}{9 x ^{2} - 16} d x

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