# Solving a differential equation using the Frobenius method

I want to solve the differential equation $$2ty'' + (1 - 2t)y' - y =0$$ using Frobenius's method. I understand the method, and I've looked up several examples; however, I can't manage to solve this problem. Here's my attempt.

Let $$y = \sum_{n=0}^{\infty} a_{n}t^{n}$$ so that $$y' = \sum_{n=0}^{\infty} na_{n}t^{n - 1}$$ and $$y'' = \sum_{n = 0}^{\infty} n(n-1) a_{n}t^{n -2}$$.

Plugging these values in, we find

$$2t \sum_{n=0}^{\infty} n(n - 1)a_{n}t^{n - 2} + (1 - 2t)\sum_{n=0}^{\infty}na_{n}t^{n - 1} - \sum_{n=0}^{\infty} a_{n}t^{n} = 0.$$

I am really not sure about how to proceed here. I tried to mess around with the indices in order to make it in a nicer form. I also tried comparing coefficients, but I got nowhere. Note that we can change the index of the second sum to $$n = 1$$ without losing any terms. Also, we can change the index of the first sum to $$n = 2$$ without losing any terms. These observations didn't get me anywhere, though.

• is not $t$ meant to be the same as $x$? – user121049 Mar 11 at 10:00
• Oops. They should be the same. – user381493 Mar 11 at 10:16
• The method isn't a straightforward power series. You need to start with $y = x^m\sum_{n=0}c_nx^n$, and find out what the lowest power $m$ is. – Dylan Mar 11 at 10:42
• Using the factorization $(2tD-1)(D-1)y=0$ one finds for $u=y'-y$ the equation $2tu'-u=0$ with solution $u=C\sqrt{t}$ and thus $$y=C_1e^t+C_2\int e^{t-s}\sqrt{s}\,ds.$$ This you can now compare to the power series solutions. – LutzL Mar 11 at 11:28

Plugging these values in, we find $$2t\sum_{n=0}^{\infty}n(n-1)a_nt^{n-2}+(1-2t)\sum_{n=0}^{\infty}na_nt^{n-1}-\sum_{n=0}^{\infty}a_nt^n = 0$$

The next step is to sort things so we have matching powers of $$t$$. For some particular $$k$$, what's the coefficient of $$t^k$$?

To do this, we first bring distribute those polynomials in $$t$$ across the sums: $$\sum_{n=0}^{\infty}2n(n-1)a_nt^{n-1}+\sum_{n=0}^{\infty}na_nt^{n-1}-\sum_{n=0}^{\infty}2na_nt^{n}-\sum_{n=0}^{\infty}a_nt^n = 0$$ Next, we shift indices so that all of the sums match up. Let $$k=n-1$$ in the first and second sums, and $$k=n$$ in the third and fourth: $$\sum_{k=-1}^{\infty}2(k+1)ka_{k+1}t^k+\sum_{k=-1}^{\infty}(k+1)a_{k+1}t^k-\sum_{k=0}^{\infty}2ka_kt^k-\sum_{k=0}^{\infty}a_kt^k = 0$$ And now that they're all lined up, we recombine the sums. They all have the same factor $$x^k$$, so $$\sum_{k=0}^{\infty}\left[2(k+1)ka_{k+1}+(k+1)a_{k+1}-2ka_k-a_k\right]t^k = 0$$ $$\sum_{k=0}^{\infty}\left[(k+1)(2k+1)a_{k+1}-(2k+1)a_k\right]t^k = 0$$ Since the $$k=-1$$ terms were both zero, we dropped them from the sum. Now, the coefficient $$(2k+1)[(k+1)a_{k+1}-a_k]$$ of $$a_k$$ is zero for each $$k\ge 0$$, so we must have $$a_k=(k+1)a_{k+1}$$. That leads to the familiar exponential series $$a_n=\frac{a_0}{n!}$$.

Now, there's a nagging problem - we only got a one-dimensional solution space. This is a second order equation - where's the other solution? What happened is that the differential equation is singular - that leading $$2t$$ coefficient of $$y''$$ is zero at $$t=0$$. There's another solution, but it doesn't have a power series at zero because it isn't differentiable there.

Can we find that other solution with this method? Well, look at what happens if we don't require the exponents $$k$$ to be integers. We can solve the system "backwards" to get ever-increasing values of $$a_k$$ as $$k\to-\infty$$, but that just leads to a divergent series. We only get a convergent series if the process stops somewhere.

When does it stop? If we're working with integer $$k$$, it stops when we hit $$k=0$$. The next step $$k=1$$ multiplies the coefficient by zero. This is, of course, the solution we've already found.
Our other chance is that $$2k+1$$ coefficient. Whatever $$a_{1/2}$$ is, $$a_{-1/2}$$ can be anything. If we set $$a_{-1/2}=0$$, then $$a_{-3/2}=0$$ and so on.

So, that's the alternative: the other solution is of the form $$y=a_{\frac12}t^{\frac12}+a_{\frac32}t^{\frac12}+\cdots+a_{n+\frac12}t^{n+\frac12}+\cdots$$ Can you find that series, too?