Question

Solve with the method of Frobenius (generalized power series): x^2y'' + xy' -9y = 0 where y = y(x)

Answer 1

ok you need to first assume that the solution is \[y=\sum_{n=0}^\infty a_nx^{n+c}\]
So \[y'=\sum_{n=0}^\infty (n+c)a_nx^{n+c-1}\]
\[y''=\sum_{n=0}^\infty (n+c)(n+c-1)a_nx^{n+c-2}\]
Substituting in D.E:

\[x^2\sum_{n=0}^\infty (n+c)(n+c-1)a_nx^{n+c-2}+x\sum_{n=0}^\infty (n+c)a_nx^{n+c-1}-9\sum_{n=0}^\infty a_nx^{n+c}=0\]
\[\sum_{n=0}^\infty (n+c)(n+c-1)a_nx^{n+c}+\sum_{n=0}^\infty (n+c)a_nx^{n+c}-\sum_{n=0}^\infty 9a_nx^{n+c}=0\]
Extracting coefficient of x^{n+c}:
\[(n+c)(n+c-1)a_n + (n+c)a_n - 9a_n = 0, n\geq0\]
\[a_n((n+c)(n+c-1)+(n+c)-9)=0\]
\[a_n((n+c)(n+c-1+1)-9)=0\]
\[a_n((n+c)^2-9)=0\]
\[a_n(n+c+3)(n+c-3)=0\]
When n=0, a_0 is never zero, so (c+3)(c-3)=0, therefore c=-3 or c=3.

(to be continued)

Answer 2

When n>0 and c=3, we get:

a_n(n+6)(n) = 0
Neither n+6 nor n can be zero (we're assuming n>0), so we get a_n=0 for all n>0.

When n>0 and c=-3, we get:

a_n(n)(n-6)=0

Hence a_n=0 except when n=6.

So when c=3, solution is \[y_1=a_0x^3\]
When c=-3, solution is \[y_2=a_0x^{-3}+a_6x^3\]

Therefore, general solution is y=y_1+y_2
\[y= Ax^{-3} + Bx^3\]