Question

(2-x^2)y'' + xy' + 16y = 0 and y'' - x^2y' - 3xy = 0 Find the general form in powers of x, stating the recurrence relation and radius of convergence.

Answer 1

should i assume this relates to using power series?

Answer 2

yes it is

Answer 3

and is it a system of equations that must be satisfied as opposed to simply 2 separate equations?

Answer 4

Sorry I posted 2 separate questions in 1 that I should have posted as 2.

Answer 5

whew!! so it is spose to be 2 separate equations .... thats a relief

Answer 6

granted it was more interesting as a system, but at leasts its simpler now :)

Answer 7

assume theres a power series that works;

y = :(0,inf): an x^n
y' = :(1,inf): an n x^(n-1)
y'' = :(2,inf): an n(n-1) x^(n-2)

and plug them into the eq

Answer 8

y'' - x^2y' - 3xy = 0



y'' = :(2,inf): an n(n-1) x^(n-2)
-x^2 y' = :(1,inf): -an n x^(n+1)
-3x y = :(0,inf): -3an x^(n+1)


we want them all to start at the same summation and exponents for good measure

Answer 9

:(2,inf): an n(n-1) x^(n-2)
:(1+3,inf): -a(n-3) (n-3) x^(n+1-3)
:(0+3,inf): -3a(n-3) x^(n+1-3)


:(2-2,inf): an n(n-1+2) x^(n-2+2)
:(4-2,inf): -a(n-3+2) (n+2-3) x^(n-2+2)
:(3-2,inf): -3a(n-3+2) x^(n-2+2)



:(0,inf): a(n+2) (n+2)(n+1) x^(n)
:(2,inf): -a(n-1) (n-1) x^(n)
:(1,inf): -3a(n-1) x^(n)


you think thats good so far?

Answer 10

2 a2 + 6x a3 + :(2,inf): a(n+2) (n+2)(n+1) x^(n)
:(2,inf): -a(n-1) (n-1) x^(n)
-3x a0 + :(2,inf): -3a(n-1) x^(n)

with any luck im remembering this correctly

Answer 11

a0,a2,a3 = 0


[a(n+2) (n+2)(n+1) -a(n-1) (n-1) -3a(n-1)] = 0


i spose i shoulda left an an in there someplace, dontcha think?

Answer 12

im going to paper for this

Answer 13

http://archive.org/details/CalculusRevisitedPart3-PowerSeriesSolutions

while im dillydallying about, this is a good review

Answer 14

:(2,inf): an n(n-1) x^(n-2)
:(1+3,inf): -a(n-3) (n-3) x^(n+1-3)
:(0+3,inf): -3a(n-3) x^(n+1-3)


:(2,inf): an n(n-1) x^(n-2)
:(4,inf): -a(n-3) (n-3) x^(n-2)
:(3,inf): -3a(n-3) x^(n-2)


2 a2 + 6 a3 x + :(4,inf): an n(n-1) x^(n-2)
:(4,inf): -a(n-3) (n-3) x^(n-2)
- 3 a0 x + :(4,inf): -3a(n-3) x^(n-2)

2 a2 = 0 when a2 = 0

6 a3 - 3 a0 = 0

:(4,inf): [ an n(n-1) - a(n-3) (n-3) -3a(n-3) ] x^(n-2) = 0

an n(n-1) = a(n-3) [ (n-3) + 3]

an = a(n-3) /(n-1)

we have a recurrsive equation such that :
\[a_2=0;\ a_0=2a_3\]
\[a_n=\frac{a_{n-3}}{n-1}\]
\[a_3=\frac{a_0}{2};\ a_6=\frac{a_0}{2*5};\ a_9=\frac{a_0}{2*5*8}...\]
\[a_4=\frac{a_1}{3};\ a_7=\frac{a_1}{3*6};\ a_{10}=\frac{a_1}{3*6*9}...\]
\[a_5=\frac{a_2}{4};\ a_8=\frac{a_2}{4*7};\ a_{11}=\frac{a_2}{4*7*11}...=0\]

which is all i think i can do for that; not real sure how to wrap it up tho