Question

Find the series solution of the differential equation:

Answer 1

\[\large (6+x^2)y''-xy'+12y=0\]
\[\large y=\sum_{n=0}^{\infty}a_nx^n,~~~y'=\sum_{n=1}^{\infty}na_nx^{n-1},~~~y''=\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}\]
After plugging these in, I got
\[\sum_{n=0}^{\infty}6(n+2)(n+1)a_{n+2}x^n+\sum_{n=2}^{\infty}n(n-1)a_nx^n-\sum_{n=1}^{\infty}na_nx^n+\sum_{n=0}^{\infty}12a_nx^n\]
I was able to simplify this into:
\[36a_3x+11a_1x+12a_2+12a_0+\]\[\sum_{n=2}^{\infty}[6(n^2+3n+2)a_{n+2}+(n^2-n)a_n-na_n+12a_n]x^n\]

Where do I go after this?

Answer 2

@ganeshie8

Answer 3

oops, I left out a n+2 in the second term exponent after I plugged them in, I believe

Answer 4

Oh wait, no, I'm being stupid because I plugged in the x^2. :P

Answer 5

after rearranging you get just below right ?

\[\sum_{n=0}^{\infty}[6(n^2+3n+2)a_{n+2}+(n^2-n)a_n-na_n+12a_n]x^n=0\]

Answer 6

Plus the terms above it

Answer 7

how did you get them ?

Answer 8

I plugged in n=0,1 to allow the indices to match without changing exponent degree

Answer 9

notice, below is true \[\sum\limits_{n=2}^5 n(n-1) =\sum\limits_{n=\color{red}{0}}^5 n(n-1) \]

Answer 10

Oh, I see. So I guess I could then match the indices to n=1 instead of n=2

Answer 11

so you don't really need to pull anything out, simply change the index and you will be fine..

Answer 12

\[\sum_{n=0}^{\infty}6(n+2)(n+1)a_{n+2}x^n+\sum_{n=2}^{\infty}n(n-1)a_nx^n-\sum_{n=1}^{\infty}na_nx^n+\sum_{n=0}^{\infty}12a_nx^n\]

is same as

\[\sum_{n=0}^{\infty}6(n+2)(n+1)a_{n+2}x^n+\sum_{n=\color{red}{0}}^{\infty}n(n-1)a_nx^n-\sum_{n=\color{red}{0}}^{\infty}na_nx^n+\sum_{n=0}^{\infty}12a_nx^n\]

Answer 13

I'm supposed to pull terms out. That's what I was taught in both the book and lecture. We have to do something with those terms to find a recursive formula that will help find the series solution.

Answer 14

Not in this problem, there is absolutely no need to pull anything out

Answer 15

Okay, then how do I proceed with finding the solution?

Answer 16

the only reason for pulling few terms out is to align the exponents,
in our case, all the exponents are aligning perfectly... so we're good.

Answer 17

Wait why is the 3rd term in your post the same for both scenarios? Wouldn't the first term be a1x in the first scenario and 0 in the second?

Answer 18

Oh but then 0+next term would be the same

Answer 19

Yes, the first term of \(\sum\limits_{n=0}^{\infty} na_n\) is 0

That's precicely the reason we don't need to pull anything out

Answer 20

aha, gotcha. So how do I find the recursive formula for this guy?

Answer 21

\(\sum\limits_{n=1}^{\infty} na_nx^n =\sum\limits_{n=\color{red}{0}}^{\infty} na_nx^n \)

Answer 22

recursive relation is
\[6(n^2+3n+2)a_{n+2} + (n^2-2n+12)a_n=0\]

Answer 23

it looks tough...

Answer 24

Okay, so then I would solve for \(a_{n+2}\) right?

Answer 25

what exactly the instructions say ?
do you really need to find the closed form or just finding the first few terms of power series is sufficient ?

Answer 26

because, solving above recurrence relation is really a pain...

Answer 27

Our professor sent us home with the problem saying solve the differential equation with series solutions. In the past homework, we had to solve for the first 4-8 terms. So I imagine that that's what we have to do..

Answer 28

then it is simple, just isolate \(a_{n+2}\) and plugin n=0,1,2,...

Answer 29

Okay, but then what's a0 and a1? Once I plug in n=0, I'll immediately start with a2

Answer 30

you may think of \(a_0\) and \(a_1\) as arbitrary constants and express all other coefficents in terms of \(a_0\) and \(a_1\)

Answer 31

Hey @CShrix I also got
\[12(a_2+a_0)+(36a_3+11a_1)x \\ +\sum_{n=2}^{\infty} (6(n+2)(n+1)a_{n+2}+n(n-1)a_n-n a_n+12a_n)x^n=0 \\ \text{ I think this means we can say } \\ a_2=a_0 \text{ and } 36a_3=-11a_1 \text{ then plug in } n=2,3,4,5,... \text{ for the other part } \\ 6(n+2)(n+1)a_{n+2}+n(n-1)a_n-n a_n+12a_n=0\]

Answer 32

oops a_2=-a_0

Answer 33

@freckles

\[12(a_2+a_0)+(36a_3+11a_1)x \\ +\sum_{n=2}^{\infty} (6(n+2)(n+1)a_{n+2}+n(n-1)a_n-n a_n+12a_n)x^n=0 \]

is same as

\[\sum_{n=\color{red}{0}}^{\infty} (6(n+2)(n+1)a_{n+2}+n(n-1)a_n-n a_n+12a_n)x^n=0 \]

right?

Answer 34

no

Answer 35

but your way also works @ganeshie8 Like I plug in 0 and 1
and also got the a_2=-a_0 and 36a_3=-11a_1 into your thingy

Answer 36

oh is it the same as is what you said

Answer 37

I think so

Answer 38

But I was just trying to let @CShrix know what he had was also right

Answer 39

Thanks @freckles !! I was scared that I was losing my mind because I thought I was wrong haha. Good to also see that there are other ways to think through the problem.

Give me a few minutes so I can check and see if my terms are correct

Answer 40

Okay, so for the first few terms, I have
\(\large a_2=-a_0\) and \(\large a_3=\frac{-11}{36}a_1\)
After plugging in values into the equation within the summation, I got:
\[\large n=2~~~~~a_4=\frac{ 1 }{ 6 }a_0\]\[\large n=3~~~~~a_5=\frac{ 55 }{ 312 }a_1\]\[\large n=4~~~~~a_6=\frac{ -1 }{ 72 }a_0\]\[\large n=5~~~~~a_7=\frac{ -165 }{ 8736 }a_1\]

Answer 41

I did not check the arithmetic but it looks good to me!

Answer 42

maybe I did something wrong but
I got
\[a_5=\frac{-a_3}{8}=\frac{11}{36(8)}a_1\]

Answer 43

also I'm very prone to arithmetic mistakes right (I'm a little sick)
or other mistakes

Answer 44

so you can double check that again

Answer 45

Oh I think I made an error for that one. You're correct! Must've done something stupid lol

Answer 46

So now that I have my terms, how can I plug it into a single combined solution?

Answer 47

you remember your solution form you chose a long time ago in the problem

Answer 48

\[y=\sum_{n=0}^{\infty} a_n x^n \\ =a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+ \cdots \]

Answer 49

Ah, haha. Duh, that makes sense. I did that on the homework but I guess I had a memory lapse X)

Thanks again to both @freckles and @ganeshie8 ! :)

Answer 50

i did check one more of your terms I got something different for a_6 too

Answer 51

Yeah, I realized the algebraic mistake on that one too after I checked a_5 :P

Answer 52

k