= Solution
The sum of the two <independent> exponential times has a <hypoexponential distribution>. By <convolution of probability densities>, for $t>0$ and $d=\lambda_1-\lambda_2>0$,
$$
\begin{aligned}
g(t)&=\int_0^t\lambda_1e^{-\lambda_1s}\lambda_2e^{-\lambda_2(t-s)}\,ds\\
&=\lambda_1\lambda_2e^{-\lambda_2t}\int_0^t e^{-ds}\,ds
=\frac{\lambda_1\lambda_2}{d}e^{-\lambda_2t}(1-e^{-dt}).
\end{aligned}
$$
Therefore
$$
\boxed{g(t)=\frac{\lambda_1\lambda_2}{\lambda_1-\lambda_2}
(e^{-\lambda_2t}-e^{-\lambda_1t}),\quad t>0,}
$$
and $g(t)=0$ for $t<0$. This is a nonnegative normalized <probability density function>. At the equal-rate boundary it tends to $\lambda^2te^{-\lambda t}$, a shape-two <Gamma distribution>.
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