Question

given the following diff equation
\[(1+x^2)y"-4xy'+6y=0\]
and \[y(x)=\sum_{0}^{\infty} a_n x^n\]
find a formula for a_n

Answer 1

define your derivatives and insert them

Answer 2

\[(1+x^2)\sum_{2}^{\infty}n(n-1)a_nx^{n-2}-4x\sum_{1}^{\infty}na_nx^n+6\sum_{0}^{\infty}a_nx^n\]

Answer 3

=0 that is

Answer 4

the top infinity is assumed

distribute now

Answer 5

i did evaluate y for n=o and n=1 to get all of the sum start at same starting point

Answer 6

you have to distrubute, then align your exponents, then align your indexes

Answer 7

x^2 i took to inside as well as x

Answer 8

indexes are aligned with evaluating the other two so they start at n=2
no?

Answer 9

(1+x^2)y" -4xy′ +6y =0

y'' +x^2y" -4xy′ +6y = 0


[a_n n(n-1) x^(n-2)]_2
+x^2[a_n n(n-1) x^(n-2)]_2
-4x[a_n n x^(n-1)]_1
+6[a_n x^n]_0 = 0


distribute your 'terms'

[a_n n(n-1) x^(n-2)]_2
+[a_n n(n-1) x^(n+2-2)]_2
+[-4a_n n x^(n+1-1)]_1
+[6a_n x^n]_0 = 0



[a_n n(n-1) x^(n-2)]_2
+[a_n n(n-1) x^n]_2
+[-4a_n n x^n]_1
+[6a_n x^n]_0 = 0


we still have an exponent that is not in line, i always adjust to the 'lowest' exponent:
n-2 < n


[a_n n(n-1) x^(n-2)]_2
+[a_{n-2} (n-2)(n-2-1) x^(n-2)]_2+2
+[-4a_{n-2} (n-2) x^(n-2)]_1+2
+[6a_{n-2} x^(n-2)]_0+2 = 0



[a_n n(n-1) x^(n-2)]_2
+[a_{n-2} (n-2)(n-2-1) x^(n-2)]_4
+[-4a_{n-2} (n-2) x^(n-2)]_3
+[6a_{n-2} x^(n-2)]_2 = 0

now we can align our indexes up to 4

Answer 10

hmm did the first line yu did let me read the last lines to see where i missed

Answer 11

srry its not in latex coding, but this amount of coding would lag my system to a halt

Answer 12

oh good i got it! that's what i needed i was doing by sub using a different index
i can take from there
since the sub didn't go well for i started evaluating to make the indices aligned

Answer 13

that's good enough for me thanks :)

Answer 14

good luck ;)

Answer 15

my prof was doing sub too, but he didn't get the answer lol

Answer 16

my efforts .... the wolf says there are i's in it so i might have messed up along the way :)

[a_n n(n-1) x^(n-2)]_2
+[a_{n-2} (n-2)(n-3) x^(n-2)]_4
+[-4a_{n-2} (n-2) x^(n-2)]_3
+[6a_{n-2} x^(n-2)]_2 = 0





a_2 2(2-1) x^(2-2)
+ a_3 3(3-1) x^(3-2)
+[a_n n(n-1) x^(n-2)]_4

+[a_{n-2} (n-2)(n-3) x^(n-2)]_4

- 4a_{3-2} (3-2) x^(3-2)
+[-4a_{n-2} (n-2) x^(n-2)]_4

+ 6a_{2-2} x^(2-2)
+ 6a_{3-2} x^(3-2)
+[6a_{n-2} x^(n-2)]_4 = 0




2a_2 + 6a_0 + 6a_3 x + 2a_1 x

+[a_n n(n-1) x^(n-2)]_4
+[a_{n-2} (n-2)(n-3) x^(n-2)]_4
+[-4a_{n-2} (n-2) x^(n-2)]_4
+[6a_{n-2} x^(n-2)]_4 = 0


now that indexes and exponents are all aligned, we can combine the summations:


2a_2 + 6a_0 + 6a_3 x + 2a_1 x

+[a_n n(n-1)
+ a_{n-2} (n-2)(n-3)
-4a_{n-2} (n-2)
+6a_{n-2}
x^(n-2)]_4 = 0


+[a_n n(n-1)
+ a_{n-2}((n-2)(n-3-4)+6)
x^(n-2)]_4 = 0

+[a_n n(n-1)
+ a_{n-2}((n-2)(n-7)+6)
x^(n-2)]_4 = 0



a_n n(n-1) + a_{n-2}((n-2)(n-7)+6) = 0

a_n n(n-1) = -a_{n-2}((n-2)(n-7)+6)

a_n = -a_{n-2} ((n-2)(n-7)+6)/(n(n-1))
-------------------------------------

2a_2 + 6a_0 = 0
a_2 = -3 a_0

6a_3 + 2a_1 = 0
a_3 = -1/3 a_1


a_2 and a_3 are defined in terms of a_0 and a_1


a_4 = -a_{2} ((2)(-3)+6)/(4(3)) ; notice a_4 = 0
a_5 = -a_{3} ((3)(-2)+6)/(5(4)) ; notice a_5 = 0
a_6 = -a_{4} ((4)(-1)+6)/(6(5))
a_7 = -a_{5} ((5)(0)+6)/(7(6))
a_8 = -a_{6} ((6)(1)+6)/(8(7))
a_9 = -a_{7} ((7)(2)+6)/(9(8))


all out terms for n>=4 drop out, they are 0

y = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + [0]

y = a_0 + a_1 x + a_0 (-3)x^2 + a_1 (-1/3)x^3

y = a_0 + a_0 (-3)x^2 + a_1 x + a_1 (-1/3)x^3

y = a_0 (1-3x^2) + a_1 x(3-x^2)/3