Is there an explicit formula for a integral recurrence relation (concerning polynomials)?

Question

Is there some general explicit solution to the equation $p_n(x)= k\int_{0}^{x} p_{n-1}(x) dx +b$, where $p(0)=a$, and $k$, $b, a$ are just constants. I need this for a more specific example, but I thought maybe there was a more general formula, though I haven't found anything after searching for a while. Is there a way to express the polynomial $p_n(x)$ explicitly?

Answer 1

After computing the first few such polynomials, I found that the pattern was

$$p_n(x)=\frac{1}{n!}a(xk)^n+b\sum_{m=0}^{n-1}\frac{1}{m!}(xk)^m$$

It is easy to show that this is the general form by induction. The statement is obviously true for $n=0$. Then

$$k\int_0^xp_n(t)dt+b=k\int_0^x\left(\frac{1}{n!}a(xk)^n+b\sum_{m=0}^{n-1}\frac{1}{m!}(xk)^m\right)dx +b$$

$$=\frac{1}{(n+1)!}a(xk)^{n+1}+b\sum_{m=0}^{n-1}\frac{1}{(m+1)!}(xk)^{m+1}+b$$

$$=\frac{1}{(n+1)!}a(xk)^{n+1}+b\sum_{m=1}^{n}\frac{1}{m!}(xk)^{m}+b$$

$$=\frac{1}{(n+1)!}a(xk)^{n+1}+b\sum_{m=0}^{n}\frac{1}{m!}(xk)^{m}=p_{n+1}(x)$$

Interestingly, as $n$ approaches infinity these polynomials approach the function $b e^{kx}$. This should come as no surprise as

$$k\int_0^x be^{kt}dt+b=kb\left(\frac{1}{k}(e^{kx}-e^0)\right)+b=be^{kx}+b-b=be^{kx}$$