What is the solution for y^'=((e^{2x+y}))/y ?

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

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y^'=((e^{2x+y}))/y

y^{^{'}} = \frac{( e ^{2 x + y} )}{y}

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

Solution steps

y^{'} (x) = \frac{( e ^{2 x + y} )}{y}

Solve separable ODE:

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

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

\frac{d}{d x} (tan (y)) = \frac{d x}{d x}

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

(1 - y^{2}) d x = - (x y - y) d y

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

5 x y \frac{d y}{d x} = 3, y > 0

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