What is the solution for (e^{x+y}+ye^y)dx+(xe^y-1)dy=0 ?

The solution for (e^{x+y}+ye^y)dx+(xe^y-1)dy=0 is y=\W(-1/(e^{(c_{1)-e^x)/x}x})+(c_{1}-e^x)/x

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(e^{x+y}+ye^y)dx+(xe^y-1)dy=0

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

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

Solution steps

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

Solve exact ODE:

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

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

\frac{s + 1}{s ^{2} + 2 s + 5}

\frac{d}{d x} (\frac{x ^{2}}{9 - x ^{2}})

\frac{\partial}{\partial x} (2 y x)

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

f (x) = \frac{( x + 1 ) ( 3 x - 7 )}{3 x + 1}

What is the solution for (e^{x+y}+ye^y)dx+(xe^y-1)dy=0 ?
The solution for (e^{x+y}+ye^y)dx+(xe^y-1)dy=0 is y=\W(-1/(e^{(c_{1)-e^x)/x}x})+(c_{1}-e^x)/x