What is the solution for y^'+x^2y=x^2y^4 ?

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

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y^'+x^2y=x^2y^4

y^{^{'}} + x^{2} y = x^{2} y^{4}

Solution steps

y^{'} (x) + x^{2} y = x^{2} y^{4}

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y^{^{'}} + \frac{x}{x ^{2} + 1} y = - 4 x

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