I'm asking again politely. I'm afraid to be blocked by the admin. being blocked is deprivation to learn from my complicated / hard problems that being stuck with by answering it by your knowledge. anyway let this thing over.
I'm solving Frobenius method exercise in number 7, I countered this:
$\ y = x^2y''+ (2x+3x^2)y' -2y $
I tried to solve the problem by indicial Equation from Frobenius method.
$\ r(r-1)+ p_0r + q_0 $
By this formula, I did this.
I let $\ p_0 = 2 $ while $\ q_0 = 0 $
$\ r^2-r +2r $
then $\ r^2+r $
$\ r(r + 1) $
the r is zero and negative one
$\ r=0, -1 $
However, I check the answer sheet at back of the book
but the answer of r is different than the one I have solved
$\ r=0, -2 $
the solution or the answer is this.
$\ y_1(x) = \frac {1}{x^2} (2-6x+9x^2), y_2(x) =\sum_{n=1}^\infty \frac{(-1)^{n-1} 3^n x^n}{(n+2+!} $
I solved the $\ y_2 (x) $ and I got it right, however my problem, I'm stuck with $\ y_1 (x) $
I try to solve $\ y_1 (x) $
$\ r = -2 $
$\ y'' = x^{n+r} (n+r)^2 - (n+r) $
$\ y' = x^{n+r} (n+r) $
$\ y = x^{n+r} $
$\ y = x^2x^{n+r} (n+r)^2 - (n+r)+ (2x+3x^2)x^{n+r} (n+r) -2 x^{n+r} $
This brings
$\ y = x^2 x^{n+r-2} (n+r)^2 - (n+r)+ (2x x^{n+r-1} (n+r) +3x^2 x^{n+r-1} (n+r)) -2 x^{n+r} $
and
$\ y = x^{n+r} (n+r)^2 - (n+r) + (2 x^{n+r} (n+r) +3x x^{n+r+1} (n+r)) -2 x^{n+r} $
$\ y = x^{n+r} (n+r)^2 - (n+r) + 2 x^{n+r} (n+r) -2 x^{n+r} + 3x^{n+r+1} (n+r) $
make $\ 3x^{n+r+1} (n+r) $ to $\ 3x^{n+r} (n+r-1) C_{n-1} $
after that
$\ y = x^{n+r} (n+r)^2 - (n+r) + 2 x^{n+r} (n+r) -2 x^{n+r} C_n + 3x^{n+r} (n+r-1)C_{n-1} $
all $\ x^{n+r} $ will be collected
$\ y = x^{n+r} [(n+r)^2 - (n+r) + 2 (n+r) -2 C_n + 3 (n+r-1)C_{n-1} ] $
focus on $\ y =(n+r)^2 - (n+r) + 2 (n+r) -2 C_n + 3 (n+r-1)C_{n-1} $
substitute $\ r = -2 $
$\ (n-2)^2 - (n-2) + 2 (n-2) -2 C_n + 3 (n-2-1)C_{n-1} $
then minus $\ - (n-2) + 2 (n-2) $ to bring this
$\ (n-2)^2 + (n-2) -2 C_n + 3 (n-3)C_{n-1} $
Find the factors of $\ (n-2)^2 + (n-2) -2 $
$\ (n) (n-3) C_n + 3 (n-3)C_{n-1} $
distribute n and 3
$\ (n^2-3n) C_n + 3n-9 C_{n-1} $
Transpose $\ 3n-9 C_{n-1}$
$\ (n^2-3n) C_n = - 3n+9 C_{n-1} $
divide by $\ (n^2-3n) $ both sides
$\ C_n = \frac{-3n+9 C_{n-1}}{(n^2-3n)} $
add $\ +n^2-n^2 $
$\ C_n = \frac{-3n+9 +n^2-n^2 C_{n-1}}{(n^2-3n)} $
I continue to solve
$\ C_n = \frac{n^2-3n C_{n-1}}{(n^2-3n)}+\frac {9-n^2}{n^2 -3n} $
I ended up like this
$\ C_n = 1C_{n-1}+\frac {9-n^2}{n^2 -3n} $
I'm stuck, I can't get the right solution can you help to help me to bridge my solution from the problem into this
$\ y_1(x) = \frac {1}{x^2} (2-6x+9x^2) $