What is the solution for 2x^2y-x^3y^'=y^3 ?

The solution for 2x^2y-x^3y^'=y^3 is y=(x^2)/(sqrt(x^2+c_{1))},y=-(x^2)/(sqrt(x^2+c_{1))}

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2x^2y-x^3y^'=y^3

2 x^{2} y - x^{3} y^{^{'}} = y^{3}

y = \frac{x ^{2}}{x ^{2} + c _{1}}, y = - \frac{x ^{2}}{x ^{2} + c _{1}}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{x ^{2}}{x ^{2} + c _{1}}, y = - \frac{x ^{2}}{x ^{2} + c _{1}}

y = \frac{x ^{2}}{x ^{2} + c _{1}}, y = - \frac{x ^{2}}{x ^{2} + c _{1}}

\int \frac{x ^{2}}{1 + x ^{6}} d x

\int \frac{x ^{2}}{a} e^{- \frac{x}{a}} d x

\frac{\partial}{\partial x} (e^{(t - x)})

\frac{d}{d x} (4 x^{2} - 5)

y = 3 x^{3} - 10 x - \frac{4}{x ^{3}}

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