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

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

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

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

y = 3 \frac{1}{W ( \frac{x ^{3}}{c _{1}} )} x

Solution steps

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

Solve using substitution:

y = 3 \frac{1}{W ( \frac{x ^{3}}{c _{1}} )} x

y = 3 \frac{1}{W ( \frac{x ^{3}}{c _{1}} )} x

\int x^{2} x^{3} + 9 d x

\int e^{c}

\frac{\partial}{\partial x} (4 x - \frac{y ^{2}}{x ^{2}})

(0, 2), (2, 0)

y = 3 sin (x), at x = \frac{π}{4}

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