# Simple Frobenius problem without recurrence relation?

I am just learning frobenius method in my 'math methods in physics' class. The first problem i am trying to solve is $$x^2y''-xy'+n^2y=0$$ (where n is a constant).

I know that i have to plug in the maclaurin expansion of y which results in me getting this summation

$$y(z) = \sum_{k = 0}^{ \infty} a_kx^{(k+s)}[n^2-(k+s)+(k+s)(k+s-1)] = 0$$ [ by asserting that $$a_0 \neq 0$$ is non-zero i solved for s and got $$s=1 \pm \sqrt{1-n^2}$$

Now i am stuck on how to continue to get the coefficients a_1, a_2, and so on. I remember seeing an example that solved for a recurrence relation but since every term had the same power of x when we plugged it back in we cant do that method.

How do i continue?

Given

$$\tag 1 x^2y''-xy'+n^2y=0$$

where n is a constant, solve the DEQ by the method of Frobenius.

Method I Euler-Cauchy

We see that this is a Euler-Cauchy equation, so lets choose:

$$y = x^r \rightarrow y' = r x^{r-1} \rightarrow y'' = r(r-1)x^{r-2}$$

Plugging this back into $(1)$, yields:

$$x^2(r)(r-1)x^{r-1} - x(r)x^{r-1} + n^2 x^r = (r(r-1) + r + n^2)x^r = 0 \rightarrow r_{1,2} = 1 \pm \sqrt{1-n^2}$$

This gives us the linear combination of a solution as:

$$y(x) = c_1 x^{1 + \sqrt{1-n^2}} + c_2x^{1 - \sqrt{1-n^2}}$$

Method II Frobenius Method

Using the Method of Frobenius, we have:

$$\tag 2 y = x^s \sum_{k=0}^\infty a_k~x_k$$

This gives us:

$$\tag 3 y' = \sum_{k=0}^\infty (k+s)~ a_k~x^{k+s-1}$$

$$\tag 4 y'' = \sum_{k=0}^\infty (k+s)(k+s-1)~a_k~x^{k+s-2}$$

Now, in order to calculate the coefficients, lets set up a table of $x$ powers for each term in the DEQ as:

$$\begin{array}{c|c|c|c|c} \text{} & x^s & x^{s+1} & x^{s+2} & \ldots & x^{k+s} \\ \hline x^2 y'' & s(s-1)~a_0 & (s+1)s~a_1 & (s+2)(s+1)~a_2 & \ldots & (k+s)(k+s-1)~a_k \\ \hline -x y' & -s~a_0 & -(s+1)~a_1 & -(s+2)~ a_2 & \ldots & -(k+s)~a_k\\ \hline n^2 y & n^2~a_0 & n^2~a_1 & n^2~a_2 & \ldots & n^2~a_k \\ \hline \end{array}$$

The total coefficient of each power of $x$ must be zero. If we sum the terms in column $x^s$ for $a_0$, we get:

$$(s^2 -2s+ n^2)a_0 = 0 \rightarrow s^2 -2s+ n^2 = 0 \rightarrow s_{1,2} = 1 \pm ~\sqrt{1-n^2}$$

This equation for $s$ is called the indicial equation and we solve it to find our values of $s$.

Next, if we sum the general column for $a_k$, we get (regardless of substituting $s_{1,2}$:

$$(k^2+2 k s-2 k+s^2-2 s + n^2)~a_k = 0 \rightarrow a_k = 0$$

Since $n$ is a constant.

So, from $(1)$, we have:

$$y(x) = x^s \sum_{n=0}^\infty a_n~x_n = a_0 x^s \rightarrow y = c_1 x^{1+\sqrt{1-n^2}} + c_2x^{1-\sqrt{1-n^2}}$$

• Thanks for the reply! Did i calculate s incorrectly? Where exactly did i go wrong? – aaron Oct 28 '13 at 4:30
• I cannot figure out what you did, but you might want to spend some time on these examples its.caltech.edu/~esp/acm95b/frobenius.pdf. Regards – Amzoti Oct 28 '13 at 4:32
• I accidently wrote the problem down wrong in the original post. The DE is -xy' not +xy'. So i believe that this method is correct but the s i calculated is correct i think. But this still causes me problems in understanding how to do the problem. I believe your methodology will work again for this new value for s, but i am getting lost trying to connect your last few steps. could you explain a bit more? Thanks! – aaron Oct 28 '13 at 4:35
• I reworked the problem. Please, when you write problems like this, double check them. This is why I stay away from these, because people make a lot of mistakes just writing the problem correctly. – Amzoti Oct 28 '13 at 5:06
• Thanks so much! and so sorry for the typo, i could have sworn i typed it correctly, i will be more careful in the future. Could you explan the step where you sum the general column a_n? I dont see how you reduced it to what you got – aaron Oct 28 '13 at 5:08