Solving $(x-1)y'' -xy' +y = 0$ using Frobenius Series

Question

I wish to solve $(x-1)y'' -xy' +y = 0$ using Frobenius series. The issue I am running into is the $x$ coefficient of the first derivative as I wish to find a solution centered at $x=1$. How would one take care of this? Can you express $x$ as a finite power series of $x = (x-1) + (x-1)^0$?

Answer 1

Let's see... we start with $z=x-1$. Then $$\frac{dy}{dx}=\frac{dy}{dz}\frac{dz}{dx}=\frac{dy}{dz}$$ And so $$\frac{d^2y}{dx^2}=\frac{d^2y}{dz^2}$$ So that $$z\frac{d^2y}{dz^2}-(z+1)\frac{dy}{dz}+y=0$$ We let $$y=\sum_{k=0}^{\infty}a_kz^{k+r}$$ Which leads to $$\sum_{k=0}^{\infty}(k+r)(k+r-1)a_kz^{k+r-1}-\sum_{k=0}^{\infty}(k+r)a_kz^{k+r}-\sum_{k=0}^{\infty}(k+r)a_kz^{k+r-1}+\sum_{k=0}^{\infty}a_kz^{k+r}=0$$ The coefficient of $a_0z^{r-1}$ is $r(r-1)-r=r(r-2)=0$, so the difference between the roots is an integer. But we catch a break: for $k\ge1$ matching coefficients of like powers of $z$ we get $$\left[(k+r+1)(k+r)-(k+r+1)\right]a_{k+1}=(k+r+1)(k+r-1)a_{k+1}=(k+r-1)a_k$$ Thus if $r=0$ and $k=1$ we find that $a_2$ is unconstrained so there is no logarithmic solution. We could start with $r=0$ so that $a_1=a_0$ and then let the unconstrained $a_2=0$. This yields the solution $y=a_0(z+1)=a_0x$. Or we could let the indeterminate $$\frac{k+r-1}{k+r-1}=1$$ even when $k=1$ and $r=0$. This leads to $$(k+1)a_{k+1}=\frac{(k+1)!}{k!}a_{k+1}=a_k$$ so $$(k+1)!a_{k+1}=k!a_k=0!a_0=a_0$$ and we have $$a_k=\frac{a_0}{k!}$$ which results in $$y=a_0\sum_{k=0}^{\infty}\frac{z^k}{k!}=a_0e^z=\frac{a_0}ee^x$$ Our overall solution is thus $$y=c_1x+c_2e^x$$ confirmed by differentiation. Now, can you edit your post to explain where you encountered a problem with all of this?