What is the solution for (y^3-2x^2y)dx+2x^3dy=0 ?

The solution for (y^3-2x^2y)dx+2x^3dy=0 is y= x/(sqrt(ln(x)+c_{1))},y=-x/(sqrt(ln(x)+c_{1))}

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(y^3-2x^2y)dx+2x^3dy=0

(y^{3} - 2 x^{2} y) d x + 2 x^{3} d y = 0

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

Solution steps

(y^{3} - 2 x^{2} y) d x + 2 x^{3} d y = 0

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}}

\int x^{3} x^{2} + 7 d x

cos (θ)

\frac{\partial}{\partial x} (x ln (y - z))

x \to 2 lim (\frac{x ^{2} - 4}{x cos ( \frac{x π}{2} ) - 2 cos ( \frac{x π}{2} )})

- 2 x^{2} y + y^{^{'}} = - 7 x^{2}

What is the solution for (y^3-2x^2y)dx+2x^3dy=0 ?
The solution for (y^3-2x^2y)dx+2x^3dy=0 is y= x/(sqrt(ln(x)+c_{1))},y=-x/(sqrt(ln(x)+c_{1))}