# Solving differential equation with power series $y'' - xy' - y = 0$

I need a work check

$$y'' -xy' - y = 0$$ given $$y(0) = 1$$ and $$y'(0) = 0$$

so we have:

$$\begin{split} y &= \sum_{n=0}^\infty C_nx^n\\ y' &= \sum_{n=1}^\infty nC_nx^{n-1}\\ y' &= \sum_{n=2}^\infty n(n-1)C_nx^{n-2} \end{split}$$

so making a substitution into the original differential equation:

$$y' = \sum_{n=2}^\infty n(n-1)C_nx^{n-2} - \sum_{n=0}^\infty nC_nx^{n-1} - \sum_{n=0}^\infty C_nx^{n} = 0$$

$$y' = \sum_{n=0}^\infty (n+2)(n+1)C_{n+2}x^{n} - \sum_{n=0}^\infty (n+1)C_nx^n = 0$$

$$C_{n+2} = \frac{C_n}{n+2}$$

so first a few terms:

$$C_0 = C_0$$ and $$C_1 = C_1$$ and $$C_2 = \frac{C_0}{2}$$ and $$C_3 = \frac{C_1}{3}$$ and $$C_4 = \frac{C_2}{4} = \frac{C_0}{4 \cdot 2}$$ and $$c_5 = \frac{c_3}{5} = \frac{C_1}{5 \cdot 3}$$

even terms are: $$\frac{C_0}{2^n \cdot n!}$$ odd terms are: $$\frac{C_1}{1 \cdot 3 \cdot 5 \cdot ...(2n+1)}$$

and so $$y = \sum_{n=0}^\infty \frac{c_0}{2^n \cdot n!}x^{2n} + \sum_{n=0}^\infty \frac{c_1}{1 \cdot 3 \cdot 5 \cdot (2n+1)} x^{2n+1}$$

Where do I go from here? if I do $$y(0)$$, I don't get 1. How do I take the derivative of this to check the initial condition $$y'(0) = 0$$

• What do you get for $y(0)$?? You get $c_0$, right? So $c_0=1$. Same idea for $y'(0)$. Differentiate term-by-term, you did it at the beginning, do it here. – GEdgar May 10 at 21:44

You're given initial conditions so you must define $$c_0$$ and $$c_1$$ to satisfy these initial conditions.

Notice that you can write your series as

$$y(x) =c_0+ \sum_{n=1}^\infty \frac{c_0}{2^n \cdot n!}x^{2n} + \sum_{n=0}^\infty \frac{c_1}{1 \cdot 3 \cdot 5 \cdot (2n+1)} x^{2n+1}$$

where I have simply taken the first term of the first series out front. Now it is clear, when $$x=0$$, both series are zero, so you are left with

$$y(0)=c_0=1$$

so $$c_0=1$$. Now, take the derivative with respect to $$x$$ of your entire function,

$$y'(x)=\sum_{n=1}^\infty \frac{2n}{2^n \cdot n!}x^{2n-1} + \sum_{n=0}^\infty (2n+1)\frac{c_1}{1 \cdot 3 \cdot 5 \cdot (2n+1)} x^{2n}$$

Now notice again when $$x$$ is zero, the only surviving term is the zeroeth term of the second series, giving you

$$y'(0)=\frac{c_1}{15}=0$$

and thus your entire series is

$$y = \sum_{n=0}^\infty \frac{1}{2^n \cdot n!}x^{2n}$$