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

The solution for (e^y-1)y^'=2+cos(x) is y=-\W(-1/(e^{c_{1)+2x+sin(x)}})-2x-sin(x)-c_{1}

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

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

y = - W (- \frac{1}{e ^{c_{1} + 2 x + s i n (x)}}) - 2 x - sin (x) - c_{1}

Solution steps

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

Solve separable ODE:

y = - W (- \frac{1}{e ^{c_{1} + 2 x + s i n (x)}}) - 2 x - sin (x) - c_{1}

y = - W (- \frac{1}{e ^{c_{1} + 2 x + s i n (x)}}) - 2 x - sin (x) - c_{1}

\frac{d y}{d x} = 8 x^{7} y

(x + 3) d y - \frac{1}{y ^{2}} d x = 0

y^{^{'}} = 4 y + 3

(1 - cos (x)) \frac{d y}{d x} = y sin (x), y = 4, x = π

\frac{e ^{x}}{e ^{2 x} - 4} d x = 5 y d y

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