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

The solution for xy^2y^'=x+x^3,y(1)=2 is y=\sqrt[3]{3x+x^3+4}

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xy^2y^'=x+x^3,y(1)=2

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

Solution steps

x y^{2} y^{'} (x) = x + x^{3}

Solve separable ODE:

\frac{d y}{d x} = \frac{1 + x}{y}

y^{^{'}} = \frac{sin ( x )}{y}, y (\frac{π}{2}) = 1

(1 - cos (x)) (\frac{d y}{d x}) = y sin (x)

y^{^{'}} (x) = \frac{4 + 3 x ^{2}}{3 y ^{2} - 12 y}, y (0) = 2

(y cos (x)) d x + (3 sin (x)) d y = 0

What is the solution for xy^2y^'=x+x^3,y(1)=2 ?
The solution for xy^2y^'=x+x^3,y(1)=2 is y=\sqrt[3]{3x+x^3+4}