Find the distribution of $Z=X+Y$ where both $X$ and $Y$ are exponentially distributed. I have a problem where, in order to solve it, I need to find the distribution of $Z$. Say that $X\sim\text{exp}(\lambda)$ and $Y\sim\text{exp}(\mu)$. I don't want to use the convolution formula but instead the mgf's. I have that
$$M_X(t)=\frac{\lambda}{\lambda-t},\quad \quad M_Y(t)=\frac{\mu}{\mu-t}.$$
Thus, 
$$M_Z(t)=M_{X+Y}(t)=M_X(t)M_Y(t)=\frac{\lambda\mu}{\lambda\mu-\lambda t-\mu t+t^2}.$$
I'm trying to find what distribution this is the mgf for. If $X$ and $Y$ were I.I.D, then it would just be an mgf for the gamma distribution. But this is not the case since I can't rewrite it in the correct form.
Is there any way to proceeed here?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{\infty}{\expo{-y/\mu} \over \mu}
\braces{\bracks{z - y > 0}{\expo{-\pars{z - y}/\lambda} \over \lambda}}\dd y}
\\[5mm] = &\
\bracks{z > 0}{\expo{-z/\lambda} \over \lambda\mu}
\int_{0}^{z}\exp\pars{{\mu - \lambda \over \lambda\mu}\,y}\dd y
\\[5mm] = &\
\bracks{z > 0}{\expo{-z/\lambda} \over \lambda\mu}
{\expo{\pars{\mu - \lambda}z/\pars{\lambda\mu}} -
1 \over \pars{\mu - \lambda}/\pars{\lambda\mu}}\dd y =
\bbx{\bracks{z > 0}{\expo{-z/\mu} - \expo{-z/\lambda} \over \mu - \lambda}}
\end{align}
