What is the solution for y^'=((-y^2))/(-(xy-x^3)) ?

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

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y^'=((-y^2))/(-(xy-x^3))

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

y = - x (- x + x^{2} - c_{1}), y = x (x + x^{2} - c_{1})

Solution steps

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

Solve exact ODE:

y = - x (- x + x^{2} - c_{1}), y = x (x + x^{2} - c_{1})

y = - x (- x + x^{2} - c_{1}), y = x (x + x^{2} - c_{1})

\frac{\partial}{\partial y} (7 sin (x y))

\frac{\partial}{\partial x} (3 x e^{5 x y})

x \to - 4 lim (x)

\int \frac{x ^{2}}{( 1 + 2 x ) ^{\frac{1}{3}}} d x

\frac{\partial}{\partial y} (e^{x y} y^{2} + \frac{x}{y})

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