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

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

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

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

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

Solution steps

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

Solve Bernoulli ODE:

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

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

x \to 0 lim (\frac{- 6}{x ^{7}})

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