Solve $(1 + x)y’ = py; y(0) = 1$, where $p$ is an arbitrary constant.

First I plugged in the guess $y = sum_{n = 0}^infty a_n x^n$:

$(1 + x)(sum_{n = 0}^infty a_n x^n)’ = psum_{n = 0}^infty a_n x^n$

Then I expanded the derivative and multiplication:

$sum_{n = 0}^infty n a_n x^{n – 1} + sum_{n = 0}^infty n a_n x^n = psum_{n = 0}^infty a_n x^n$

Then I shifted the left index (the first term yielding $0$ allows the lower bound to remain $0$) and algebraically combined the summations:

$sum_{n = 0}^infty (n + 1)a_{n + 1} x^n + (n – p)a_n x^n = 0$

This leads to the following recurrence relation:

$a_{n + 1} = frac{p – n}{n + 1}a_n$

Thus for various values of $n$:

$a_1 = p a_0$, $a_2 = frac{p(p – 1)}{2}a_0$, $a_3 = frac{p(p – 1)(p – 2)}{6} a_0$, etc.

So applying definitions for the exponential taylor series and falling factorial, the guessed solution would be:

$y = sum_{n = 0}^infty frac{p! a_0 x^n}{n! (p – n)!} = sum_{n = 0}^infty a_0 e^x p^{underline n}$

Solving the initial value problem:

$1 = sum_{n = 0}^infty a_0 e^0 p^{underline n} implies a_0 = frac{1}{sum_{n = 0}^infty p^{underline n}}$

My final solution is:

$y = frac{sum_{n = 0}^infty e^x p^{underline n}}{sum_{n = 0}^infty p^{underline n}}$

However, the answer is supposed to be $y = (1 + x)^p$. Are these identical, or did I make an error somewhere?