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?
