What is the solution for 4y^'+2x+2ce^{2y}y^'=0 ?

The solution for 4y^'+2x+2ce^{2y}y^'=0 is y=-(\W(c/(2e^{(-2c_{1)+x^2)/2}}))/(2)-(x^2)/4+(c_{1})/2

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4y^'+2x+2ce^{2y}y^'=0

4 y^{^{'}} + 2 x + 2 c e^{2 y} y^{^{'}} = 0

y = - \frac{W ( \frac{c}{2 e ^{\frac{- 2 c _{1} + x ^{2}}{2}}} )}{2} - \frac{x ^{2}}{4} + \frac{c _{1}}{2}

Solution steps

4 y^{'} (x) + 2 x + 2 c e^{2 y} y^{'} (x) = 0

Solve separable ODE:

y = - \frac{W ( \frac{c}{2 e ^{\frac{- 2 c _{1} + x ^{2}}{2}}} )}{2} - \frac{x ^{2}}{4} + \frac{c _{1}}{2}

y = - \frac{W ( \frac{c}{2 e ^{\frac{- 2 c _{1} + x ^{2}}{2}}} )}{2} - \frac{x ^{2}}{4} + \frac{c _{1}}{2}

What is the solution for 4y^'+2x+2ce^{2y}y^'=0 ?
The solution for 4y^'+2x+2ce^{2y}y^'=0 is y=-(\W(c/(2e^{(-2c_{1)+x^2)/2}}))/(2)-(x^2)/4+(c_{1})/2