<br />
\mu'(t) + 9 \mu(t) = 15<br />
is an inhomogeneous 1st order linear ODE. The integrating factor is:
<br />
A(t) = \exp \left(\int{9 \, dt} \right) = e^{9 t}<br />
Then, we have:
<br />
\frac{d}{dt} \left( e^{9 t} \, \mu(t) \right) = 15 \, e^{9 t}<br />
Integrate once:
<br />
e^{9 t} \, \mu(t) = \frac{15}{9} e^{9 t} + C_1<br />
where C_1 is an arbitrary integration constant. We have a general solution:
<br />
\mu(t) = \frac{5}{3} + C_1 \, e^{-9 t}<br />
To find C_1, we need an inital conditon. Look at the integral equation:
<br />
\mu_x(t) = 2 + \int_{0}^{t}{\left(15 - 9 \, \mu_x(s) \right) \, ds}<br />
Substitute t = 0. The integral is zero because the upper and lower bound coincide! We have:
<br />
\mu_x(0) =2<br />
From here, we have:
<br />
2 = \frac{5}{3} + C_1 \Rightarrow C_1 = \frac{1}{3}<br />
Thus, the mean is:
<br />
\mu_x(t) = E[X_t] = \frac{5 + e^{-9 t}}{3}<br />
