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

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

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

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

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

Solution steps

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

\frac{d}{d x} (x \cdot ln (x) - x)

y = 3 x + 2 cos (x)

y = x^{4} - 3 x^{2} + 8, at x = 1

\int (\frac{1}{x ^{5}} - x^{5} - \frac{1}{9}) d x

x \to - 2 - lim (- 4 x - 8 + 3)

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