Question

Hi everyone! Can someone check my work for this infinite series solution to the DE y"=xy
I am supposed to find y1 and y2. Any advice would be welcome. Thanks! :o)

Answer 1

At first I was concerned about not having the values of Co and C1, but somehow just following y=Co+C1x+C2x^2 etc, it seemed to work itself out...If anyone cares to elaborate on how it did that , that would be great! :o)

Answer 2

so your trying power series solution eh

Answer 3

yes...ugh! :o)

Answer 4

y = sum0 an x^n


xy = sum0 an x^{n+1}
y'' = sum2 an n(n-1) x^{n-2}

line up the powers, i work it to the lowest, so n-2


xy = sum0+3 a[n-3] x^{n-3+1}
y'' = sum2 an n(n-1) x^{n-2}


xy = sum3 a[n-3] x^{n-2}
y'' = sum2 an n(n-1) x^{n-2}

line up the indexes

xy = sum3 a[n-3] x^{n-2}
y'' = 2a[2] + sum3 an n(n-1) x^{n-2}

combine the an coeefs since it all lines up

y'' - xy = 2a[2] + sum3 {an n(n-1) - a[n-3]} x^{n-2} = 0

we good to here?

Answer 5

did you look at my work?

Answer 6

a0 = a0
a1 = a1
a2 = 0

i did, but i tend to have to work this out to make sure it flows ... lets see if we end up the same

Answer 7

it's much easier to refer to what I already did instead of trying to figure out all the typing and syntax

Answer 8

im up to you on the bottom of the first page

Answer 9

gotcha! :o)
just so you know though, it's a little hard for me to follow the typing fyi

Answer 10

our rule for the coeffs

an n(n-1) - a[n-3] = 0

a[n] = a[n-3]/(n(n-1))

\[a_n=\frac{a_{n-3}}{n(n-1)}~:~n\ge3\]
we have the same setup so far

Answer 11

why do we need 4 non-negative terms?

Answer 12

because of the instructions

Answer 13

and it tends to be best if you can adjust your denominators to be factorials ... or at least thats the way i work them

Answer 14

i think I accidentally did 5...opps

Answer 15

well some of the denominators are not factorials though...some are like:
9.8.6.5.3.2 they are not 9.8.7.6.5.4.3.2.1 etc they skip so I can't write as factorials unless you know something that I don't

Answer 16

you can make them factorials by multiply by some useful form of 1 to fill in what they need

\[a_n=\frac{a_{n-3}}{n(n-1)}~:~n\ge3\]
\[a_3=\frac{a_{0}}{3(2)}=a_0\frac{1}{3!}\]
\[a_4=\frac{a_{1}}{4(3)}=a_1\frac{2}{4!}\]
\[a_6=\frac{a_{3}}{6(5)}=a_0\frac{1(4)}{6!}\]
\[a_7=\frac{a_{4}}{7(6)}=a_1\frac{2(5)}{7!}\]
\[a_9=\frac{a_{6}}{9(8)}=a_0\frac{1(4)(7)}{9!}\]
\[a_10=\frac{a_{7}}{10(9)}=a_1\frac{2(5)(8)}{10!}\]

Answer 17

our arbitrary constants are ...well you have c's. c0 and c1

Answer 18

\[y_1=a_{3n}=a_0\frac{\prod\limits_{k=1}^{n}(3k-2)}{(3n)!}\]
\[y_1=a_{3n+1}=a_1\frac{\prod\limits_{k=1}^{n}(3k-1)}{(3n+1)!}\]

Answer 19

you as long as all your computations on the coeffs are right, it looks like you did a fine job

Answer 20

my teacher doesn't want the answers in series form, just y1 and y2 expanded

Answer 21

you did fine ...

Answer 22

okay great! :o)
do you mind if I ask a couple things about your method?

Answer 23

sure

Answer 24

i mean go ahead and ask lol ...

Answer 25

you typed in that n>=3 thingy
my teacher did that too but I didn't know why
I thought I heard him say he just switched back the index to n= instead of k= because he said it's easier to calculate
do you think I heard him correctly?
I mean, is that why you changed your index back as well?

Answer 26

ive tried it both ways, and im more comfortable (learned it that way) defining a_n= instead of a_(n+3)=

when we drop the exponents down to the 'lowest' value, the n-a to me reads as the lowest value of a is the largest number.

that allows me to move my index up instead of down. its a habit.

now because when i lined all the powers and indexes up, my summation started at 3, thats why i defined n to start at 3

a_(n-3) after all only makes sense if we start with n=3

Answer 27

ok, so my teacher(or you for that matter) totally didn't need to switch back the indices from k= back to n= then...it's simply a function of your comfort level and preference then right?

Answer 28

correct

Answer 29

okay, well I better stick with k= I think, the less switching for me the better...algebra hates me! :o~

Answer 30

:) potato tomato eh

Answer 31

lol...okay something else...

Answer 32

I completely understand why you would want to represent the denominators with factorials so that you can represent the y1 and y2 as power series representations...

so...

taking my C9 on page 2 for example, how on Earth do I do that?
C9=Co/9.8.6.5.3.2

Answer 33

you are missing 1 and 4 and 7 underneath (the 1 is useless i know but habit)

multiply by 4(7)/4(7)

Answer 34

gimme a sec to write that out

Answer 35

so 28Co/9! ?

Answer 36

Co 28/9! yes

Answer 37

yikes! looks scary

Answer 38

spose we had something like:

1/(7.3.2) we are missing 6.5.4

1/(7.3.2) = (6.5.4)/(7.6.5.4.3.2) = 120/7!

Answer 39

yeah I think I see that now...
I was just noticing that my next missing term in the denominator would have been C10, so if I needed 3 terms in the denominator to make a factorial, I could have just done:
Co/9.8.6.5.3.2 times (4/7)(7/4)(1/10)(10/1) = 280Co/10! OR...
Co/9.8.6.5.3.2 times (4/7)(7/4)(10/10)= 280Co/10!
Is that correct? :o/

Answer 40

the thing is, by filling in the factorials, we notice a pattern of products forming up top for a given nth term

1
1(1+3)
1(1+3)(1+3+3)
1(1+3)(1+3+3)(1+3+3+3)

product (k=1 to n) of: 1 + 3(k-1)

or

product (k=1 to n) of: 3k - 2 for some nth term

Answer 41

\[c_{10}=c_1\frac{1}{10.9.7.6.4.3}=c_1\frac{\color{red}{8}.\color{red}{5}.\color{red}{2}}{10.9.\color{red}{8}.7.6.\color{red}{5}.4.3.\color{red}{2}}\]
\[~~~~~~~~~~~~\implies c_1\frac{80}{10!}\]

Answer 42

yes, I should have known it wasn't that simple, I never did the math to figure out C10 but regardless, I get it!
A more simple approach would be this:
"Hey there Ms. math student, are you missing 3 6 and 8 from the denominator? No problem, just multiply the quantity by (3/3)(6/6)(8/8)"

Right? :o)

Answer 43

correct :)

Answer 44

lol...last thing if you don't mind...

Answer 45

at the beginning you stated right away that Co=Co and C1=C1
now for me I was a bit worried just from lack of experience that I didn't have values for Co and C1, but am I correct in assuming that as long as I put everything in terms of Co and C1 and then factor out at the end Co and C1, I never never actually need to state that Co=Co and C1=C1 do I?

Answer 46

before you learnt power series, you dealt with arbitrary constants ....

the C0 and C1 are nothing more than the arbitrary constants. If we had an initial value we might even be able to solve for them.

Answer 47

the wolf gives us the solution as
\[y=c_0Ai(x)+c_1Bi(x)\]
Ai is the Airy function
Bi is the Airy Bi function

i got no idea what those are, but they are what they call the power series that we generated.

Answer 48

so then I don't need to worry about them since I am guided by:
y=Co+C1x+C2x^2 etc
and I put all the C2 AND c3'S etc in terms of C2 and C3 etc
they get factored out, leaving your y1 and y2, and hence no need to be a scared-E pants right? :o)

Answer 49

sorry...typo

Answer 50

no need to be concerned :)

Answer 51

so then I don't need to worry about them since I am guided by:
y=Co+C1x+C2x^2 etc
and I put all the C2 AND c3'S etc in terms of Co and C1 etc
they get factored out, leaving your y1 and y2, and hence no need to be a scared-E pants right? :o)

Answer 52

I think I got it finally!
Any further advice for me?

Answer 53

dont sweat the small stuff ....

Answer 54

:o)
Thanks sooooo much @amistre64 !
You definitely get a medal!
They should have a cookie button on here as well! nomnomnom :o)
Thanks!

Answer 55

>(B)< ; butterscotches lol

Answer 56

omg...butterscotch topping on french vanilla ice cream! mmm

Answer 57

I think I need a break now! whew
no ice cream but I do have M and M's meh! lol
Have a great night!

Answer 58

imma leave before one of us goes into a diabetic coma ... good luck ;)

Answer 59

Thanks you too