Difference of two exponential distribution Let $X\sim Exp(\lambda_1)$ and $Y\sim Exp(\lambda_2)$, with $Z = X - Y$.
I am trying to find the pdf of Z, i.e. $f_Z(z)$. 
Here is what I have got:
\begin{align*}
        f_Z(z)  &= \int_0^{z}f_X(z+y)f_Y(y)dy\\
                &= \int_0^{z}\lambda_1 e^{-\lambda_1(z+y)}\lambda_2 e^{-\lambda_2y}dy\\
                &= \lambda_1\lambda_2e^{-\lambda_1z}\int_0^{z}e^{-(\lambda_1+\lambda_2)y}dy\\
                &= \lambda_1\lambda_2e^{-\lambda_1z}\bigg[\frac{e^{-(\lambda_1+\lambda_2)y}}{-(\lambda_1+\lambda_2)}\bigg]_{y=0}^{y=z}\\
                &= \frac{\lambda_1\lambda_2e^{-\lambda_1z}}{\lambda_1+\lambda_2}(1 - e^{-{(\lambda_1 + \lambda_2)}z})\\
                &= \frac{\lambda_1\lambda_2}{\lambda_1+\lambda_2}(e^{-\lambda_1z} - e^{-{(2\lambda_1 + \lambda_2)}z})
    \end{align*}
But I have looked at other page from mathstackexchange, pdf of the difference of two exponentially distributed random variables, but I cannot get that answer.
Edit:
$X$ and $Y$ are independent.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\mrm{f}_{Z}\pars{z}  & =
\int_{0}^{\infty}\mrm{f}_{X}\pars{z + y}\mrm{f}_{Y}\pars{y}
\bracks{z + y > 0}\dd y
\\[5mm] & =
\int_{0}^{\infty}\lambda_{1}\expo{-\lambda_{1}\pars{z + y}}
\lambda_{2}\expo{-\lambda_{2}y}\bracks{y > -z}\dd y
\\[5mm] & =
\lambda_{1}\lambda_{2}\expo{-\lambda_{1}z}
\int_{0}^{\infty}\expo{-\pars{\lambda_{1} + \lambda_{2}}y}
\bracks{y > - z}\dd y
\\[8mm] & =
\lambda_{1}\lambda_{2}\expo{-\lambda_{1}z}\left\{%
\bracks{z < 0}\int_{-z}^{\infty}\expo{-\pars{\lambda_{1} + \lambda_{2}}y}\dd y\right.
\\[2mm] & 
\phantom{=
\lambda_{1}\lambda_{2}\expo{-\lambda_{1}z}\,\,\,}
\left.\mbox{} + \bracks{z > 0}\int_{0}^{\infty}
\expo{-\pars{\lambda_{1} + \lambda_{2}}y}\dd y\right\}
\\[8mm] & =
\lambda_{1}\lambda_{2}\expo{-\lambda_{1}z}\bracks{%
\bracks{z < 0}{\expo{\pars{\lambda_{1} + \lambda_{2}}z} \over
\lambda_{1} + \lambda_{2}} +
\bracks{z > 0}{1 \over \lambda_{1} + \lambda_{2}}}
\\[5mm] & =
\bbx{{\lambda_{1}\lambda_{2} \over \lambda_{1} + \lambda_{2}}\braces{\bracks{z < 0}\,\expo{\lambda_{2}z} +
\bracks{z > 0}\expo{-\lambda_{1}z}}}
\end{align}
