63.3k views
1 vote
Need help with indicial equation and ODEs

Need help with indicial equation and ODEs-example-1
User Albano
by
8.1k points

1 Answer

0 votes
With


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

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)

the singular ODE can be written as


\displaystyle\sum_(n\ge0)\bigg[2(n+r)(n+r-1)+3(n+r)-1\bigg]a_nx^(n+r)+\sum_(n\ge0)\bigg[-2(n+r)(n+r-1)-3(n+r)\bigg]a_nx^(n+r-1)=0

The first term of the second series admits the indicial equation. When
n=0, we have


(-2r(r-1)-3r)a_0x^(r-1)=0\iff-2r^2-r=0\iffr^2+\frac12r=0

Factoring reveals two distinct roots at
-r(2r+1)=0\implies r_1=0,r_2=-\frac12 (in your case, swap
r_1 and
r_2 before submitting).

Next, shift the index of the first sum so that it starts at
n=1 by replacing
n\mapsto n-1, then consolidate the sums to get


\displaystyle\sum_(n\ge1)\bigg[2(n+r-1)(n+r-2)+3(n+r-1)-1-2(n+r)(n+r-1)-3(n+r)\bigg]a_nx^(n+r-1)

=\displaystyle x^r\sum_(n\ge1)\bigg[-4n-4r\bigg]a_nx^(n-1)

Setting
r=-\frac12, we then have this as


=x^(-1/2)\left(-2a_1-6a_2x-10a_3x^2-14a_4x^3-18a_5x^4+\cdots\right)

=x^(-1/2)\displaystyle\sum_(n\ge1)(2-4n)a_nx^(n-1)

However I don't see the connection to the given answer... It seems some information is missing, specifically about how the coefficients
a_n are related.
User Ymagine First
by
8.5k points