Consider the following differential equation on $[0,+\infty)$
$$y'(t) + \frac{1}{t-g(t)}y(t) = \frac{1}{t-g(t)} \text{ and } y(0)=0. \qquad(1)$$
First the constant function equal to $1$ is a solution and one only needs to study the following differential equation on $[0,+\infty)$
$$y'(t) + \frac{1}{t-g(t)}y(t) = 0.$$
Suppose $g(t) = (1-a)\ln(1+t)-a$, where $0<a<1$.
Is there an explicit formula for the solutions of this equations ? If not, what can be said, even numerically ?
My ultimate goal is to compute the expectation of a random variable with CDF a solution of $(1)$.
If $y'+fy=0$ then $y'/y=-f$ so $(\ln y)'=-f$ so $\ln y =-\int f +c$ and $y = e^{-\int f +c} =Ce^{-\int f} $.