Density of $X-Y$ where $X,Y$ are independent random variables with common PDF $f(x) = e^{-x}$? 
$X,Y$ are independent random variables with common PDF $f(x) = e^{-x}$ then density of $X-Y = \text{?}$

I thought of this let $ Y_1 = X + Y$, $Y_2 = \frac{X-Y}{X+Y}$, solving which gives me $X = \frac{Y_1(1 + Y_2)}{2}$, $Y = \frac{Y_1-Y_2}{2}$
then I calculated the Jacobian $J = \begin{bmatrix} \frac{1+y_2}{2} & \frac{y_1}{2} \\ \frac{1}{2} & -\frac{1}{2} \end{bmatrix}$ so that $\left|\det(J)\right| = \frac{1+y_1+y_2}{4}$
and the joint density of $Y_1,Y_2$ is the following $W(Y_1,Y_2) = \left|\det(J)\right| e^{-(y_1+y_2)}$ when $y_1,y_2> 0$ and $0$ otherwise.
Next I thought of recovering $X-Y$ as the marginal but I got stuck. I think i messed up in the variables.
Any help is great!.

I already posted an answer involving no integrals of functions of more than one variable; here's another approach.
\begin{align}
\text{First assume } u >0. \text{ Then} \\
\Pr( X-Y > u) & = \int_0^\infty \left( \int_{y+u}^\infty f_{X,Y} (x,y) \, dx \right) \,dy \\[10pt]
& = \int_0^\infty \left( \int_{y+u}^\infty e^{-x} e^{-y} \, dx \right) \,dy \\[10pt]
& = \int_0^\infty \left( e^{-y} \int_{y+u}^\infty e^{-x} \, dx \right) \,dy \\
& \qquad\text{(This can be done because $e^{-y}$ does not change as $x$ goes from something to $\infty$.)}
\\[10pt]
& = \int_0^\infty e^{-y} \cdot e^{-(y+ u)} \, dy \\[10pt]
& = \frac 1 2 e^{-u}.
\end{align}
That works if $u>0.$ Then use the fact that $Y-X$ has the same probability distribution as $X-Y$ to conclude that if $u<0$ then $\Pr(X-Y<u) = \frac 1 2 e^{u}.$
Therefore if $u>0$ then $\Pr(X-Y\le u) = 1- \dfrac 1 2 e^{-u}$ and mutatis mutandis if $u<0,$ so we get $\displaystyle f_{X-Y}(u) = \frac 1 2 e^{-|u|}.$
