What is the solution for x^2+2x-1-4yy^'=0 ?

The solution for x^2+2x-1-4yy^'=0 is y=sqrt((x^3+3(x^2-x-c_{1)))/6},y=-sqrt((x^3+3(x^2-x-c_{1)))/6}

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x^2+2x-1-4yy^'=0

x^{2} + 2 x - 1 - 4 y y^{^{'}} = 0

y = \frac{x ^{3} + 3 ( x ^{2} - x - c _{1} )}{6}, y = - \frac{x ^{3} + 3 ( x ^{2} - x - c _{1} )}{6}

Solution steps

x^{2} + 2 x - 1 - 4 y y^{'} (x) = 0

Solve separable ODE:

y = \frac{x ^{3} + 3 ( x ^{2} - x - c _{1} )}{6}, y = - \frac{x ^{3} + 3 ( x ^{2} - x - c _{1} )}{6}

y = \frac{x ^{3} + 3 ( x ^{2} - x - c _{1} )}{6}, y = - \frac{x ^{3} + 3 ( x ^{2} - x - c _{1} )}{6}

What is the solution for x^2+2x-1-4yy^'=0 ?
The solution for x^2+2x-1-4yy^'=0 is y=sqrt((x^3+3(x^2-x-c_{1)))/6},y=-sqrt((x^3+3(x^2-x-c_{1)))/6}