What is the solution for ydx-(2y^4+2x)dy=0 ?

The solution for ydx-(2y^4+2x)dy=0 is

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ydx-(2y^4+2x)dy=0

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

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

Solution steps

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

Solve exact ODE:

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

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

x^{4} - 8 x^{2} + 2

\frac{\partial}{\partial x} (4 y^{3} sin (3 x))

x \to \infty lim (ln (8))

x \to 1 lim (- \frac{x ^{2} - 1}{x - 1})

\int \frac{1}{x 4 x ^{4} - 1} d x

What is the solution for ydx-(2y^4+2x)dy=0 ?
The solution for ydx-(2y^4+2x)dy=0 is