Q: $x^2y^{''}-(x^2+2)y=0$ [1]

Solving using frobenius

$y=\sum_{n=0}^{\infty}a_nx^{x+r}$ [2]

$ y'=\sum_{n=0}^{\infty}(n+r)a_nx^{x+r-1}$ [3]

$ y''=\sum_{n=0}^{\infty}(n+r)(n+r-1)a_nx^{x+r-2}$ [4]

inserting [2,4] into [1]

$x^2 \sum_{n=0}^{\infty}(n+r)(n+r-1)a_nx^{x+r-2}-(x^2+2)\sum_{n=0}^{\infty}a_nx^{x+r}$ [5]

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

$x^2(\sum_{n=0}^{\infty}[(n+r-1)(n+r)a_n -2a_n]x^{n}-\sum_{n=0}^{\infty}a_nx^{n+2})=0$ [6]

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



$\sum_{k=0}^{\infty}[(k+1)(k+2)a_n -2a_k]x^k-\sum_{k=2}^{\infty}a_{k-2}x^k=0$ [9]

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 [9] 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:

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


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.

  • $\begingroup$ Without initial values you should have $2$ arbitrary constants $\endgroup$
    – Raffaele
    Oct 31, 2017 at 20:06
  • $\begingroup$ I am not understanding by what you mean by initial values $\endgroup$ Oct 31, 2017 at 20:25
  • $\begingroup$ @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. $\endgroup$
    – epi163sqrt
    Nov 1, 2017 at 13:58

1 Answer 1


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*}


  • 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 $k=1$.

From (2) we obtain the expected relations: \begin{align*} [x^1]:\qquad &&4a_1&=0&\qquad &\color{blue}{a_1=0}\\ [x^2]:\qquad &&10a_2-a_0&=0&\qquad &\color{blue}{a_2=\frac{a_0}{10}}\\ [x^3]:\qquad && 18a_3-a_1&=0&\qquad &\color{blue}{a_3=0}\\ [x^4]:\qquad &&28a_4-a_2&=0&\qquad &\color{blue}{a_4=\frac{a_2}{28}=\frac{a_0}{280}} \end{align*}


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