Compute the renewal function when the interarrival distribution $F$ is such that $1-F(t) = pe^{-\mu_1t} + (1-p)e^{-\mu_2t}$ As described in title, the problem is to compute the renewal function with interarrival distribution $F$,
$$
1-F(t) = pe^{-\mu_1t} + (1-p)e^{-\mu_2t}
$$
I tried to compute renewal function M(t) with renewal equation,
$$
M(t) = F(t) + \int_0^t M(t-y) {\rm d}F(y) 
$$
but failed to solve the equation.
I also tried to solve the problem by compound Poisson process, and let $\Lambda$
follows a Bernoulli distribution, that is
$P(\Lambda = \mu_1) = p, P(\Lambda = \mu_2) = 1 - p$.
When $\Lambda$ is given, $N(t) | \Lambda$ is a Poisson process with ratio $\Lambda$, but also got stuck.
Any hint is appreciated.
 A: Let's rewrite the renewal equation as
$$M(t) = F(t) + \int_0^t M(t-y) f(y){\rm d}y$$
with
$$f(t)=F'(t)=\mu_1 p e^{-\mu_1 t}+\mu_2 (1-p) e^{-\mu_2 t} \;\; .$$
Then, using Laplace tranforms
$$\tilde{M}(s)=\int_0^{\infty}e^{-st}M(t)dt$$
$$\tilde{F}(s) = \frac{1}{s}-\frac{p}{s+\mu_1}-\frac{1-p}{s+\mu_2}$$
$$\tilde{f}(s) = \frac{\mu_1p}{s+\mu_1}+\frac{\mu_2(1-p)}{s+\mu_2}$$
and applying the Laplace transform on the renewal equation
$$\tilde{M}(s)= \frac{\tilde{F}(s)}{1-\tilde{f}(s)}$$
which after filling in the obtained and rearranging can be written as
$$\tilde{M}(s) = \frac{[\mu_1 p + \mu_2 (1-p)]s + \mu_1\mu_2}{s^2(s+\mu_1 (1-p) + \mu_2 p)}$$
Now, it is just a matter of identifying the inverse Laplace transforms which can be done by using the tables on the wikipage I quoted.
Introducing $k_1=\mu_1 p + \mu_2 (1-p)$ and $k_2=\mu_1 (1-p) + \mu_2 p$, we should have
$$M(t)=\frac{k_1}{k_2}(1-e^{-k_2t})+\frac{\mu_1\mu_2}{k_2}t - \frac{\mu_1\mu_2}{k_2^2}(1-e^{-k_2t})$$
provided I did not do any computational mistakes.
