Im clueless on how to solve the following question...
$xe^y\frac {dy}{dx} = e^y +1$
What i've done is...
$\frac {dy}{dx} = \frac 1x + \frac {1}{xe^e}; \frac {dy}{dx} - \frac {1}{xe^e} = \frac 1x $
Find the integrating factor..
$v(x) = e^{P(x)}; where P(x) = \int p(x)dx \Rightarrow P(x) = \int \frac 1x dx = ln|x| \\v(x) = e^{P(x)} = e^{ln|x|} = x; \\ y = \frac {1}{v(x)} \int v(x)q(x) dx = \frac 1x \int {x}{\frac 1x} dx = 1+c$
I know I made a mistake somewhere. Would someone advice me on this??