Power Series Solutions and Radius of Convergence for y'' + xy = 0

[Ted123]

Homework Statement

Find 2 independent solutions which are power series in x of $$y'' + xy =0$$ and find the radius of convergence of each solution.

The Attempt at a Solution

$$\sum_{n=2}^{\infty} n(n-1)a_n x^{n-2} + x\sum_{n=0}^{\infty}a_n x^n = 0$$

$$\sum_{n=-1}^{\infty} (n+3)(n+2) a_{n+3} x^{n+1} + \sum_{n=0}^{\infty} a_n x^{n+1} = 0$$

$$2a_2 + \sum_{n=0}^{\infty} [(n+3)(n+2) a_{n+3} + a_n ]x^{n+1} = 0$$

$$\implies a_2 = 0$$
$$a_{n+3} = -\frac{a_n}{(n+3)(n+2)}$$ for $$n=0,1,2,...$$

Is this right so far?

Then how to solve for $$a_n$$ to find 2 power series solutions...?

 

[tiny-tim]

HI Ted123!

Looks ok so far.

Now try it starting with a₀ = 1, and again with a₁ = 1.

 

[Ted123]

HI Ted123!

Looks ok so far.

Now try it starting with a₀ = 1, and again with a₁ = 1.

By starting the recurrence off with $$a_0 =0,a_1=1 \; :$$
$$\displaystyle a_3=0 ,\;\;\; a_4 = -\frac{a_1}{(4)(3)} = -\frac{1}{12} ,\;\;\; a_5 = 0= a_6 , \;\;\; a_7 = -\frac{a_4}{(7)(6)} = \frac{1}{504}$$
$$a_{3k+1}=\frac{(-1)^k}{3^kk!\prod\limits_{j=1}^k (3j+1)}$$
So $$\displaystyle \sum_{k=0}^{\infty} \frac{(-1)^k}{3^kk!\prod\limits_{j=1}^k (3j+1)} x^{3k+1}$$ is one solution.

Starting the recurrence off with $$a_0 = 1,a_1=0 \; :$$
$$\displaystyle a_3 = -\frac{a_0}{(3)(2)} = -\frac{1}{6} ,\;\;\; a_4 = 0 = a_5 , \;\;\; a_6 = -\frac{a_3}{(6)(5)} = \frac{1}{180}, \;\;\; a_7 = 0 = a_8 , \;\;\; a_9 = -\frac{a_6}{(9)(8)} = -\frac{1}{12960}$$
How do I work out a closed formula for $$a_{3k}$$
so that $$\sum_{k=0}^{\infty} a_{3k} x^{3k}$$ is another solution?

It definitely involves a $$(-1)^k.$$

 

[vela]

By starting the recurrence off with $$a_0 =0,a_1=1 \; :$$
$$\displaystyle a_3=0 ,\;\;\; a_4 = -\frac{a_1}{(4)(3)} = -\frac{1}{12} ,\;\;\; a_5 = 0= a_6 , \;\;\; a_7 = -\frac{a_4}{(7)(6)} = \frac{1}{504}$$
$$a_{3k+1}=\frac{(-1)^k}{3^kk!\prod\limits_{j=1}^k (3j+1)}$$

You can come up with an expression that's a little bit simpler if you note that

\begin{align}
a_1 &= 1 \\
a_4 &= -\frac{1}{3\cdot 4} = -\frac{2}{2 \cdot 3 \cdot 4} \\
a_7 &= \frac{1}{3\cdot 4\cdot 6 \cdot 7} = \frac{2 \cdot 5}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6\cdot 7} \\
a_{10} &= \frac{1}{3\cdot 4\cdot 6 \cdot 7 \cdot 9 \cdot 10} = \frac{2 \cdot 5 \cdot 8}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10} \\

\vdots
\end{align}

In any case, what's stopping you from doing the same thing you did earlier to find an expression for $a_{3k}$?

 

[Ted123]

You can come up with an expression that's a little bit simpler if you note that

\begin{align}
a_1 &= 1 \\
a_4 &= -\frac{1}{3\cdot 4} = -\frac{2}{2 \cdot 3 \cdot 4} \\
a_7 &= \frac{1}{3\cdot 4\cdot 6 \cdot 7} = \frac{2 \cdot 5}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6\cdot 7} \\
a_{10} &= \frac{1}{3\cdot 4\cdot 6 \cdot 7 \cdot 9 \cdot 10} = \frac{2 \cdot 5 \cdot 8}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10} \\

\vdots
\end{align}

In any case, what's stopping you from doing the same thing you did earlier to find an expression for $a_{3k}$?

It's OK - I've managed to find a similar closed formula for a_{3k} ... (I had a silly algebraic mistake)