Solving this recursive differential equation, knowing the first term (maybe using Laplace Transform) Function $f_n(t)$ is defined for all positive integer $n$ and is continuously differential in $t$ for $t\in[0,T]$.
We have a recursive differential equation that we would like to solve.
$$\frac{\partial f_n(t)}{\partial t}-\alpha\left(f_n(t)-f_{n-1}(t)\right)+\alpha\beta=0,$$
with the terminal condition $f_n(T)=0$.
Now we know that for $n=1$, we have $f_1(t)=\beta-\beta e^{-\alpha(T-t)}$. So we would like to find $f_n(t)$ for $n\ge 2$.
The first thought I had was solving using the integrating factor method. Doing that, we get this solution
$$f_n(t)=\int_{t}^{T}\left(\alpha\beta + \alpha f_{n-1}(s)\right)e^{-\alpha\left(s - t\right)}ds.$$
This is nice but for higher $n$, solving the integral is very messy.
Is there another way to find a closed-form expression for $f_n(t)$?
I got a hint on using the Laplace Transform, the closed-form expression of $f_n(t)$ is something like $k + (a+bx+cx^2+…) e^{-\alpha x}$ where $x=T-t$.
So using this hint, I started with $n=2$ but I don't get this form. I can compute $f_n(t)$ for each $n$ but not a closed-form for any $n$.
Could someone help please?
 A: I see a way using a "characteristic function approach".
I mean by that, the use an auxiliary variable, say $x$, generating a formal series obeying a "big" differential equation gathering all the "small" differential equations. Let us be more precise.
Let us multiply the $n$-th differential equation by $x^n$ and sum up all of them in this way:
$$\sum_{n=1}^{\infty}\frac{\partial f_n(t)}{\partial t}x^n-\alpha \sum_{n=1}^{\infty}f_n(t)x^n+\alpha \sum_{n=1}^{\infty} f_{n-1}(t)x^n+\alpha\beta \sum_{n=1}^{\infty}x^n=0 \tag{1}$$
Let us use the following notation
$$\Phi(t,x):=\sum_{n=1}^{\infty}f_n(t)x^n\tag{2}$$
Using (2), (1) becomes:
$$\frac{\partial \Phi(t,x)}{\partial t}-\alpha \Phi(t,x)+\alpha x(\Phi(t,x)\color{red}{-f_1(t)})+\alpha \beta\frac{x}{1-x}=0 \tag{3}$$
(please note the handling (in red) of the first term). Otherwise said:
$$\frac{\partial \Phi(t,x)}{\partial t}+\alpha(x-1)\Phi(t,x)=\alpha x f_1(t)-\alpha \beta\frac{x}{1-x} \tag{4}$$
with (as indicated)
$$f_1(t)=\beta-\beta e^{-\alpha(T-t)}\tag{5}$$
giving:
$$\frac{\partial \Phi(t,x)}{\partial t}+\alpha(x-1)\Phi(t,x)=\alpha \beta\frac{x^2}{1-x}-\alpha \beta x e^{-\alpha(T-t)} \tag{6}$$
(6) is an ordinary first order differential equation in variable $t$ (with $x$ considered as a parameter, i.e., a constant).
This constant coefficients differential equation can be solved using Laplace Transform. Here is the how, denoting by $P$ the Laplace Transform of $\Phi$, with customary variable $s$ in the Laplace domain:
$$s P(s,x)-\alpha(x-1) P(s,x)=\alpha \beta \left(\dfrac{s^3}{1-s^3}-xe^{-\alpha T}\dfrac{1}{s-\alpha}\right)$$
$$P(s,x)=\dfrac{\alpha \beta}{s-\alpha(x-1)} \left(\dfrac{s^3}{1-s^3}-xe^{-\alpha T}\dfrac{1}{s-\alpha}\right)$$
I haven't done the computations of the inverse Laplace Transform of $P$, in order to obtain a closed form expression for $\Phi$ but as they involve rational fractions, they are standard (ask Wolfram Alpha for example).
Having obtained $\Phi$, it will remain to expand it as a formal series in $x^n$ and "harvest" its coefficients which are the $f_n(t)$.
