What is the solution for y=(2y^4+2x)y^' ?

The solution for y=(2y^4+2x)y^' is

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y=(2y^4+2x)y^'

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

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

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}

\frac{e ^{2 x}}{sin ( 3 x )}

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\int_{0}^{1} x 3 - x^{2} d x

f (x) = [1 + e^{x}] [ln (x)], at x = 1