I am trying to solve this differential equation, I have already obtained the general solution of the given equation in parametric form, however I do not know where to get the particular solution y = -1,Could you tell me if I am doing something wrong?
The equation $$ dy/dx=(dy/dx-1)e^{dy/dx}$$ We make $$ dy/dx=p$$ Then $$y=(p-1)e^{p}$$ $$⇒dy/dp=e^{p}+(p-1)(e^{p})$$ $$⇒dy/dp=e^{p}+(p-1)(e^{p})$$ $$⇒dy/dp=e^{p}+e^{p}-e^{p}$$ $$⇒dy/dp=e^{p}$$ $$⇒dy=e^{p}p dp$$ $$⇔pdx=e^{p} p dp$$ $$⇒dx=e^{p}dp$$ $$\int 1 dx=\int e^{p} dp$$ $$⇒x=e^{p}+C$$ The general solution is $$x=e^{p}+C , y=(p-1)e^{p} $$