Question

Solve this differential equation using power series and indicial roots about (0,0):

3xy'' + 6y' + y = 0

Answer 1

Let
`y=\displaystyle\sum_(k\ge0)a_kx^k`, so that


`y'=\displaystyle\sum_(k\ge1)ka_kx^(k-1)`

`y''=\displaystyle\sum_(k\ge2)k(k-1)a_kx^(k-2)`

Substituting into the ODE gives


`\displaystyle3x\sum_(k\ge2)k(k-1)a_kx^(k-2)+6\sum_(k\ge1)ka_kx^(k-1)+\sum_(k\ge0)a_kx^k=0`

`\displaystyle3\sum_(k\ge2)k(k-1)a_kx^(k-1)+6\sum_(k\ge1)ka_kx^(k-1)+\sum_(k\ge0)a_kx^k=0`

The first series starts with a linear term, while the other two start with a constant. Extract the first term from each of the latter two series:


`\displaystyle6\sum_(k\ge1)ka_kx^(k-1)=6\sum_(k\ge2)ka_kx^(k-1)+6a_1`

`\displaystyle\sum_(k\ge0)a_kx^k=\sum_(k\ge1)a_kx^k+a_0`

Finally, to get the series to start at the same index, shift the index of the first two series by replacing
`k` with
`k+1`. Then the ODE becomes


`\displaystyle3\sum_(k\ge1)k(k+1)a_(k+1)x^k+6\sum_(k\ge1)(k+1)a_(k+1)x^k+\sum_(k\ge1)a_kx^k+6a_1+a_0=0`

which can be consolidated to get


`\displaystyle\sum_(k\ge1)\bigg[(3k(k+1)+6(k+1))a_(k+1)+a_k\bigg]x^k+6a_1+a_0=0`

`\displaystyle\sum_(k\ge1)\bigg[3(k+1)(k+2)a_(k+1)+a_k\bigg]x^k+6a_1+a_0=0`

You're fixing the solution so that it contains the origin, which means


`y(0)=\displaystyle\sum_(k\ge0)a_kx^k=a_0=0`

which in turn means
`a_1=0`. With the given recurrence, it follows that
`a_k=0` for all
`k\ge2`, so the solution would be
`y=0`. This is to be expected, since
`x=0` is clearly a singular point for the ODE.