What is the solution for ye^ydy=(1+e^{-2x})/(e^x)dx ?

The solution for ye^ydy=(1+e^{-2x})/(e^x)dx is y=\W((-3/(e^x)-1/(e^{3x)}+c_{1})/(3e))+1

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ye^ydy=(1+e^{-2x})/(e^x)dx

y e^{y} d y = \frac{1 + e ^{- 2 x}}{e ^{x}} d x

y = W (\frac{- \frac{3}{e ^{x}} - \frac{1}{e ^{3 x}} + c _{1}}{3 e}) + 1

Solution steps

y e^{y} d y = \frac{1 + e ^{- 2 x}}{e ^{x}} d x

Solve separable ODE:

y = W (\frac{- \frac{3}{e ^{x}} - \frac{1}{e ^{3 x}} + c _{1}}{3 e}) + 1

y = W (\frac{- \frac{3}{e ^{x}} - \frac{1}{e ^{3 x}} + c _{1}}{3 e}) + 1

y^{^{'}} sec (x) = 7 y^{2}, y (\frac{9 π}{2}) = \frac{1}{4}

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

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

\frac{d y}{d x} = \frac{1}{y}

\frac{d}{d x} (sin (x) - cos (y) - 2) = 0

What is the solution for ye^ydy=(1+e^{-2x})/(e^x)dx ?
The solution for ye^ydy=(1+e^{-2x})/(e^x)dx is y=\W((-3/(e^x)-1/(e^{3x)}+c_{1})/(3e))+1