# Solving $y'' - 2xy' - 2y = 0$ using power series.

$$(1) : y'' - 2xy' - 2y = 0$$

1. Determine the power series solutions to the equation (1).

2. Let $$y$$ be a a solution to the equation $$(1)$$, defined on $$\mathbb{R}$$ and from $$C^2$$, prove that for all $$x \in \mathbb{R}$$

$$\left( e^{x^2} \left( e^{-x^2} y(x) \right)' \right)' = 0$$

1. Deduce the set of solutions of the equation (1).

$$y$$ as a power series is: $$y = \sum_{n = 0}^{+ \infty} a_n x^n$$

Pluggins it in the equation $$(1)$$, we get:

$$\sum_{n = 2}^{+ \infty} \left[ (n+1)(n+2)a_{n+2} - 2na_n - 2a_n \right]x^{n} - 2a_0 - 2a_1x = 0$$

We get from this the relation:

$$a_{n+2} = \frac{2}{n} a_n$$

I do not know how to proceed to get the full solution as a power series.

as I don't know how to answer question $$2$$ and $$3$$.

• You seem to think that, if $$y = \sum_{n = 0}^{+ \infty} a_n x^n$$ then $$y''=\sum_{n = 2}^{+ \infty} (n+1)(n+2)a_{n+2}x^n$$ Not so. – Did Feb 4 at 19:14

Part 1

Too long for comment: In power-series form, the ODE is

$$\sum_{n\ge2}a_nn(n-1)x^{n-2}-2x\sum_{n\ge1}a_nnx^{n-1}-2\sum_{n\ge0}a_nx^n=0$$ $$\sum_{n\ge2}a_nn(n-1)x^{n-2}-2\sum_{n\ge1}a_nnx^n-2\sum_{n\ge0}a_nx^n=0$$

Notice the degree of the leading terms in each sum are $$0$$, $$1$$, and $$0$$. Removing the first term from the first and third sums, then consolidating the remaining sums gives

$$2a_2-2a_0+\sum_{n\ge3}a_nn(n-1)x^{n-2}-2\sum_{n\ge1}a_nnx^n-2\sum_{n\ge1}a_nx^n=0$$ $$2a_2-2a_0+\sum_{n\ge1}a_{n+2}(n+2)(n+1)x^n-2\sum_{n\ge1}a_nnx^n-2\sum_{n\ge1}a_nx^n=0$$ $$2a_2-2a_0+\sum_{n\ge1}\bigg(a_{n+2}(n+2)(n+1)-2a_n(n+1)\bigg)x^n=0\quad\color{red}{(1)}$$

(your error occurs at some point during the above steps)

$$\implies a_{n+2}=\frac{2a_n}{n+2}\quad\color{blue}{(2)}$$

Part 2

Hint: Compute the derivative on the left hand side. Some terms will vanish and you'll end up with the original ODE.

Part 3

Equation $$\color{blue}{(2)}$$ says

$$a_4=\frac{2a_2}4=\frac{2^1a_0}{2^1(2)}$$ $$a_6=\frac{2a_4}6=\frac{2^2a_0}{2^2(2\cdot3)}$$ $$a_8=\frac{2a_6}8=\frac{2^3a_0}{2^3(2\cdot3\cdot4)}$$

and so on, which suggests a pattern for the even-indexed coefficients,

$$a_{2k}=\frac{a_0}{k!}\text{ for }k\ge2$$

Then a solution to the ODE would be

$$a_0+a_0x^2+a_0\sum_{k\ge2}\frac{x^{2k}}{k!}$$

You can do a similar analysis to describe the odd-indexed coefficients. Starting with a fixed $$a_1$$, we have

$$a_3=\frac{2a_1}3$$ $$a_5=\frac{2a_3}5=\frac{2^2a_1}{3\cdot5}$$ $$a_7=\frac{2a_5}7=\frac{2^3a_1}{3\cdot5\cdot7}$$ $$\implies a_{2k-1}=\frac{2^{k-1}a_1}{3\cdot5\cdot7\cdot\cdots\cdot(2k-1)}\text{ for }k\ge1$$

and there is another solution

$$a_1\sum_{k\ge1}\frac{2^{k-1}x^{2k-1}}{\prod\limits_{1\le i\le k}(2i-1)}$$

From here I imagine you are supposed to find closed forms for these power series. The even-indexed series you should recognize right away. The second requires a bit of massaging to get something that looks familiar.

• Ah, I see where my mistake was. Do you have an indea about the other questions? Thank you. – Zouhair El Yaagoubi Feb 4 at 20:09
• Determining these 2 solutions are the set of solutions? Thank you. – Zouhair El Yaagoubi Feb 4 at 21:43