What is the solution for (2e^y-x)y^'=1 ?

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

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(2e^y-x)y^'=1

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

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

Solution steps

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

Solve exact ODE:

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

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

\int \frac{x ^{2}}{( 225 + x ^{2} ) ^{2}} d x

\int \frac{1 + x}{x - 2 x + 2 x ^{2}} d x

\int \frac{4}{4 x ^{2} - 4 x + 5} d x

x \to 0 lim (2 x + x)

\frac{8}{r ^{3}}

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