Non-linear ODE $xy' - y = e^{y'}$

Question

Could not found appropriate method to solve this non-linear first order diff. equation. $$ xy'-y = e^{y'} $$

Thanks for help

Answer 1

Let $p = \frac{dy}{dx}$. Therefore the given differential equation is, $$xp-y=e^{p}$$ Differentiating both sides with respect to $x$ gives, $$p+xp'-p=e^{p}p'$$ $$\therefore p'=0 \implies p=C$$ When we plug this in our initial expression, we get $y=Cx-e^C$ The other solution is, $$x=e^p \implies p = \ln x$$ Plugging this in our initial expression, $$y=x\ln x-x$$ which is the singular solution. Hope this helps.

Answer 2

$$xy'-y = e^{y'}$$ $$y=xy'- e^{y'}$$ It's of the form: $$y=xy'+f({y'})$$ This is Clairaut's differential equation.