# Frobenius method series solution

Q: $x^2y^{''}-(x^2+2)y=0$ 

Solving using frobenius

$y=\sum_{n=0}^{\infty}a_nx^{x+r}$ 

$y'=\sum_{n=0}^{\infty}(n+r)a_nx^{x+r-1}$ 

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

inserting [2,4] into 

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

Multiplying through but the cofficent I get

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

setting n=0 I solve for r to get the values of $r=2 , r=-1$

Now this is where my confusion comes in

so for r=2 I rearnage the series like so

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

So now what I did I want all x's to be of the same power so I did the following:

$k=n+2$

$k-2=n$

$\sum_{k=0}^{\infty}[(k+1)(k+2)a_n -2a_k]x^k-\sum_{k=2}^{\infty}a_{k-2}x^k=0$ 

expanding on the series $\sum_{k=0}^{\infty}[(k+1)(k+2)a_n - 2a_n]x^k$ so I can get the staring point of the series in the same postion I did the following

$k=0, 2a_0-2a_0=0 \rightarrow a_0=0$

$k=1, 6a_{1}-2a_{1} \rightarrow a_1=0$

$a_0=a_1=0$ is because  is equal to $0$

so now I have the equation

$\sum_{k=2}^{\infty}[(k+1)(k+2)a_n - 2a_n-a_{k-2}]x^k=0$

So I now have a recurrence relation of:

$(k+1)(k+2)a_n - 2a_n-a_{k-2}=0$

$a_n=\frac{a_{k-2}}{k(k+3)}$

Now in my book solution it has the solution of

$a_2=\frac{a_0}{2\cdot 5}$ $a_3=0$ $a_4=\frac{a_0}{2\cdot 5\cdot 4\cdot 7}$

but how is it $a_0$? I know I have gone somewhere wrong but just cant see where.

• Without initial values you should have $2$ arbitrary constants Oct 31, 2017 at 20:06
• I am not understanding by what you mean by initial values Oct 31, 2017 at 20:25
• @JasonTaylor: There are some typos in mixing up $x$ with $n$ e.g. $x^{x+r}$ instead of $x^{n+r}$ and mixing up $n$ with $k$. You might want to check the question and correct it by yourself. This might be in connection with the mistake I've addressed in my answer. I have the impression that your understanding is quite good, but there are a lot of slips of the pen which may cause sometimes problems. Nov 1, 2017 at 13:58

Your approach is fine. There is just a small mistake which causes the problem. In the following we use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series.

Let's consider again the series equation \begin{align*} \sum_{k=0}^\infty[(k+1)(k+2)a_k-2a_k]x^k-\sum_{k=2}^\infty a_{k-2}x^k=0\tag{1} \end{align*}

We consider the coefficient of $x^0$ \begin{align*} [x^0]:\qquad 2a_0-2a_0=0\qquad \text{resp.}\qquad 0=0 \end{align*}

We observe this equation provides no information at all. So $k=0$ can't be usefully used (this was the mistake) and $a_0$ is left unspecified. This becomes plausible, when we rearrange the equation (1) somewhat. We obtain \begin{align*} \sum_{k=0}^\infty&[(k+1)(k+2)a_k-2a_k]x^k-\sum_{k=2}^\infty a_{k-2}x^k\\ &=\sum_{k=0}^\infty(k+3)ka_kx^k-\sum_{k=2}^\infty a_{k-2}x^k\tag{1}\\ &=\sum_{k=1}^\infty(k+3)ka_kx^k-\sum_{k=2}^\infty a_{k-2}x^k\tag{2}\\ &=0 \end{align*}

Comment:

• In (1) we simplify $(k+1)(k+2)a_k-2a_k$ and observe the coefficient $a_0$ vanishes in the left-hand series since $(k+3)k=0$ if $k=0$.

• In (2) we can therefore start the left series with $k=1$.