Question

Find the first three nonzero terms in each of two solutions (noy1t multiples of each other) xy''+y'-y=0

I know y1 and a_n(r) but I don't know how to differntiate a_n(r).

As I know, a_n(r)= 1/(r+1)...(r+n)

Answer 1

It looks like you're trying to solve the ODE via Frobenius' method. It's not immediately clear where the solution is to be centered, but I think we can safely assume it's to be taken about
`x=0`. Dividing through by
`x`, we get


`y''+\frac{y'}x-\frac yx=0`

and we observe that
`x=0` is a regular singular point.

Now, taking a solution of the form


`y=\displaystyle\sum_(n\ge0)a_nx^(n+r)`

`\implies y'=\displaystyle\sum_(n\ge0)(n+r)a_nx^(n+r-1)`
`\implies y''=\displaystyle\sum_(n\ge0)(n+r)(n+r-1)a_nx^(n+r-2)`

and substituting into the ODE gives


`\displaystyle\sum_(n\ge0)(n+r)(n+r-1)a_nx^(n+r-1)+\sum_(n\ge0)(n+r)a_nx^(n+r-1)-\sum_(n\ge0)a_nx^(n+r)=0`

`\displaystyle\sum_(n\ge0)(n+r)^2a_nx^(n+r-1)-\sum_(n\ge0)a_nx^(n+r)=0`

`\displaystyle r^2a_0x^(r-1)+\sum_(n\ge1)(n+r)^2a_nx^(n+r-1)-\sum_(n\ge0)a_nx^(n+r)=0`

`\displaystyle r^2a_0x^(r-1)+\sum_(n\ge1)(n+r)^2a_nx^(n+r-1)-\sum_(n\ge1)a_(n-1)x^(n+r-1)=0`

`\displaystyle r^2a_0x^(r-1)+\sum_(n\ge1)\bigg((n+r)^2a_n-a_(n-1)\bigg)x^(n+r-1)=0`

The indicial polynomial admits only one root (of multiplicity two) at
`r=0`, so in fact a regular power series solution will exist. So in fact the ODE above reduces to


`\displaystyle\sum_(n\ge1)\bigg(n^2a_n-a_(n-1)\bigg)x^(n-1)=\sum_(n\ge0)\bigg((n+1)^2a_(n+1)-a_n\bigg)x^n=0`

leaving us with the recurrence


`a_(n+1)=(a_n)/((n+1)^2)`

We can solve this by recursively substituting the right hand side:


`a_n=(a_(n-1))/(n^2)=(a_(n-2))/(n^2(n-1)^2)=\cdots=(a_0)/((n!)^2)`

and so one of the solutions will be


`y_1(x)=\displaystyle a_0\sum_(n\ge0)(x^n)/((n!)^2)`

Now the literature on the Frobenius method when the indicial polynomial has repeated roots suggests a second solution will take the form


`y_2(x)=Cy_1\ln x+\displaystyle\sum_(n\ge0)b_nx^n`

but it's not clear to me *why* this works.