Question

Xy''+2y'-xy by frobenius method

Answer 1

First note that
`x=0` is a regular singular point; in particular
`x=0` is a pole of order 1 for
`\frac2x`.

We seek a solution of the form


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

where
`r` is to be determined. Differentiating, we have


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

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

and substituting into the ODE gives


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

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

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

`\displaystyle r(r+1)a_0x^(r-1)+(r+1)(r+2)a_1x^r+\sum_(n\ge2)(n+r)(n+r+1)a_nx^(n+r-1)-\sum_(n\ge0)a_nx^(n+r+1)=0`

`\displaystyle r(r+1)a_0x^(r-1)+(r+1)(r+2)a_1x^r+\sum_(n\ge2)(n+r)(n+r+1)a_nx^(n+r-1)-\sum_(n\ge2)a_(n-2)x^(n+r-1)=0`

`\displaystyle r(r+1)a_0x^(r-1)+(r+1)(r+2)a_1x^r+\sum_(n\ge2)\bigg((n+r)(n+r+1)a_n-a_(n-2)\bigg)x^(n+r-1)=0`

The indicial polynomial,
`r(r+1)`, has roots at
`r=0` and
`r=-1`. Because these roots are separated by an integer, we have to be a bit more careful, but we'll get back to this later.

When
`r=0`, we have the recurrence


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

valid for
`n\ge2`. When
`n=2k`, with
`k\in\{0,1,2,3,\ldots\}`, we find


`a_0=a_0`

`a_2=(a_0)/(3\cdot2)=(a_0)/(3!)`

`a_4=(a_2)/(5\cdot4)=(a_0)/(5!)`

`a_6=(a_4)/(7\cdot6)=(a_0)/(7!)`

and so on, with a general pattern of


`a_(n=2k)=(a_0)/((2k+1)!)`

Similarly, when
`n=2k+1` for
`k\in\{0,1,2,3,\ldots\}`, we find


`a_1=a_1`

`a_3=(a_1)/(4\cdot3)=(2a_1)/(4!)`

`a_5=(a_3)/(6\cdot5)=(2a_1)/(6!)`

`a_7=(a_5)/(8\cdot7)=(2a_1)/(8!)`

and so on, with the general pattern


`a_(n=2k+1)=(2a_1)/((2k+2)!)`

So the first indicial root admits the solution


`y=\displaystyle a_0\sum_(k\ge0)(x^(2k))/((2k+1)!)+a_1\sum_(k\ge0)(x^(2k+1))/((2k+2)!)`

`y=\displaystyle \frac{a_0}x\sum_(k\ge0)(x^(2k+1))/((2k+1)!)+\frac{a_1}x\sum_(k\ge0)(x^(2k+2))/((2k+2)!)`

`y=\displaystyle \frac{a_0}x\sum_(k\ge0)(x^(2k+1))/((2k+1)!)+\frac{a_1}x\sum_(k\ge0)(x^(2k+2))/((2k+2)!)`

which you can recognize as the power series for
`\frac{\sinh x}x` and
`\frac{\cosh x}x`.

To be more precise, the second series actually converges to
`\frac{\cosh x-1}x`, which doesn't satisfy the ODE. However, remember that the indicial equation had two roots that differed by a constant. When
`r=-1`, we may seek a second solution of the form


`y=cy_1\ln x+x^(-1)\displaystyle\sum_(n\ge0)b_nx^n`

where
`y_1=\frac{\sinh x+\cosh x-1}x`. Substituting this into the ODE, you'll find that
`c=0`, and so we're left with


`y=x^(-1)\displaystyle\sum_(n\ge0)b_nx^n`

`y=\frac{b_0}x+b_1+b_2x+b_3x^2+\cdots`

Expanding
`y_1`, you'll see that all the terms
`x^n` with
`n\ge0` in the expansion of this new solutions are already accounted for, so this new solution really only adds one fundamental solution of the form
`y_2=\frac1x`. Adding this to
`y_1`, we end up with just
`\frac{\sinh x+\cosh x}x`.

This means the general solution for the ODE is


`y=C_1\frac{\sinh x}x+C_2\frac{\cosh x}x`