# Differential Equation Solution By Power Series

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?