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

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

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(2x-y)+(2y-x)y^'=0

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

y = \frac{x + - 3 x ^{2} + c _{1}}{2}, y = \frac{x - - 3 x ^{2} + c _{1}}{2}

Solution steps

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

Solve exact ODE:

y = \frac{x + - 3 x ^{2} + c _{1}}{2}, y = \frac{x - - 3 x ^{2} + c _{1}}{2}

y = \frac{x + - 3 x ^{2} + c _{1}}{2}, y = \frac{x - - 3 x ^{2} + c _{1}}{2}

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