Question

Consider [math] x=0 [/math] as a regular singular point of the differential equation [math] x^2y''+xy'+(x^2-\frac{4}{9})y=0 [/math]. Find the indicial roots of the singularity using the general form of the indicial equation [math] r(r-1)+a_0+b_0=0 [/math] which is applicable to the standard form of a second order differential equation [math] y''+P(x)y'+Q(x)y=0 [/math] when [math] xP(x)=a_0+a_1x+a_2x^2+... [/math] and [math] x^2Q(x)=b_0+b_1x+b_2x^2+... [/math]

Answer 1

omg nobody can understand your equations written like that it'll take 10 minutes to decipher

Answer 2

\[ x ^{2}y" + xy' + (x ^{2}- 4/9) = 0,\]
Bessel equation of order 2/3, looking for infinite series solution using the Frobenius method.

Answer 3

\[y = \sum_{n=0}^{\infty}(a _{n}x ^{n + r})\]
gets substituted into the given equation to get, after simplifying,
\[(r ^{2} - 4/9)a _{0} + ((r+1)^{2} - 4/9) a _{1}\]
\[+\sum_{n=2}^{\infty}(((n+r)^{2}-4/9)a _{0} + a _{n-2})=0.\]

Answer 4

Set \[ r ^{2} - 4/9 = 0 \]

Answer 5

Substitute r = 2/3 into the summation part to get the recursive equation
\[n(n+4/3)a _{n} = -a _{n-2},\]
n>=2

Answer 6

This gives one of the two linearly independent solutions to the Bessel equation.

Answer 7

I believe you can fill in the rest of the details to get the final form of the first solution. Can you get the other?