What is the solution for 2yy^'=2x^3-1 ?

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

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2yy^'=2x^3-1

2 y y^{^{'}} = 2 x^{3} - 1

y = \frac{x ^{4}}{2} - x + c_{1}, y = - \frac{x ^{4}}{2} - x + c_{1}

Solution steps

2 y y^{'} (x) = 2 x^{3} - 1

Solve separable ODE:

y = \frac{x ^{4}}{2} - x + c_{1}, y = - \frac{x ^{4}}{2} - x + c_{1}

y = \frac{x ^{4}}{2} - x + c_{1}, y = - \frac{x ^{4}}{2} - x + c_{1}

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