Q:

How to solve the differential equation $e^{(y^{'}-x)}y^{'}+1=0$

How to solve this differential equation 

$e^{(y^{'}-x)}y^{'}+1=0;y(0)=0;y(1)=\frac{1}{2}$.

I am finding it very difficult to solve .Is there any specific way available for this?

A:

$$e^{y'-x}y'+1=0$$
$$y'e^{y'}=-e^x$$
The condition $y(0)=0$ implies $y'e^{y'}=-1$ which is impossible for $y'$ real.
So there is no real solution if the condition $y(0)=0$ is required.
The condition $y(1)=\frac{1}{2}$ implies $y'e^{y'}=-e$ which is impossible for $y'$ real.
Again, there is no real solution if the condition $y(1)=\frac{1}{2}$ is required.
Analytical solving, thanks to the Lambert W fonction $W(X)e^{W(X)}=X$ :
$$y'=W(-e^x)$$
$X=-e^x$ hense $dx=-e^x dX$ then $dx=\frac{dX}{X}$
$$y=\int W(-e^x)dx +C=-\int \frac{W(X)}{X}dX+C=\frac{1}{2}W(X)\left(W(X)+2\right)+C$$
$$y=\frac{1}{2}W(-e^x)\left(W(-e^x)+2\right)+C$$
$W(-e^x)$ is real only if $x\leq -1$

