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

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

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

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

y = \frac{x}{2 ln ( x ) + c _{1}}, y = - \frac{x}{2 ln ( x ) + c _{1}}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{x}{2 ln ( x ) + c _{1}}, y = - \frac{x}{2 ln ( x ) + c _{1}}

y = \frac{x}{2 ln ( x ) + c _{1}}, y = - \frac{x}{2 ln ( x ) + c _{1}}

\int cos^{3} (\frac{x}{13}) d x

x \to 0 lim (x + 6)

y = (9 x^{4} + 5 x)^{3}

\int e^{8 x + 7} d x

\frac{d x}{d t} = b (1 - x) x

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