Question

Calc 3 / sort of differential equations question
Solve using a power series:
I am given y′′ − (x^2 + 8)y′ + y = 0 centered around x0= 3
I've taken the derivatives and plugged them into the original ODE. Where I'm stuck is simplifying the second term (which is more of a Calc 3 question I think): -[(x2 + 8)] multiplied by the sum from zero to infinity of na(sub n) (x-3)^n-1

Answer 1

Hint :
\[x^2+8 = (x-3)^2 + 6(x-3)+17\]

Answer 2

I'm assuming I need to solve for x? But what do I do once I get x?
thanks :)

Answer 3

Nope, just replace \(x^2+8\) by \((x-3)^2+6(x-3)+17\)

Answer 4

ohhhhhh that makes sense. so I can plug in the (x-3)^2 and just change the value of n, what about the rest of that equation though?

Answer 5

\(y = \sum\limits_{n=0}^{\infty}a_n(x-3)^n\)

\(y' = \sum\limits_{n=1}^{\infty}a_n(n-1)(x-3)^{n-1}\)

\(y'' = \sum\limits_{n=2}^{\infty}a_n(n-1)(n-2)(x-3)^{n-2}\)

Answer 6

Plugging them in the given eqn gives :
\[\sum\limits_{n=2}^{\infty}a_n(n-1)(n-2)(x-3)^{n-2}\\-(x^2+2) \sum\limits_{n=1}^{\infty}a_n(n-1)(x-3)^{n-1} \\+ \sum\limits_{n=1}^{\infty}a_n(x-3)^{n}\\=0\]

Answer 7

Next replace \(x^2+8\) by \((x-3)^2+6(x-3)+17\)

Answer 8

Wait so where did the \[-(x ^{2} + 2)\] come from? Or was that a typo and should have been an 8?

Answer 9

sry its a typo, supposed to be x^2+8 :

\[\sum\limits_{n=2}^{\infty}a_n(n-1)(n-2)(x-3)^{n-2}\\
-(x^2+\color{red}{8}) \sum\limits_{n=1}^{\infty}a_n(n-1)(x-3)^{n-1} \\
+ \sum\limits_{n=1}^{\infty}a_n(x-3)^{n}\\
=0\]

Answer 10

okay cool that makes sense, so then for the middle term I understand how to move (x-3)^2 inside of the series, how do I move the rest inside?

Answer 11

\[\sum\limits_{n=2}^{\infty}a_n(n-1)(n-2)(x-3)^{n-2}\\
-[(x-3)^2+6(x-3)+17] \sum\limits_{n=1}^{\infty}a_n(n-1)(x-3)^{n-1} \\
+ \sum\limits_{n=1}^{\infty}a_n(x-3)^{n}\\
=0\]

Answer 12

\[\sum\limits_{n=2}^{\infty}a_n(n-1)(n-2)(x-3)^{n-2}\\
- \sum\limits_{n=1}^{\infty}a_n(n-1)(x-3)^{n+1} -6 \sum\limits_{n=1}^{\infty}a_n(n-1)(x-3)^{n}-17 \sum\limits_{n=1}^{\infty}a_n(n-1)(x-3)^{n-1}\\
+ \sum\limits_{n=1}^{\infty}a_n(x-3)^{n}\\
=0\]

Answer 13

Notice that you have 5 series in above eqn

Answer 14

They all have different exponents :
\((x-3)^{n-2},~~(x-3)^{n+1},~~(x-3)^n,~~(x-3)^{n-1}\)

Answer 15

Your next step is to make them all have same exponent : \((x-3)^n\)

Answer 16

Do you know how to do that ?

Answer 17

Is that where I change the lower boundary?

Answer 18

Yes

Answer 19

simply use substitution

Answer 20

Let me show you how to do it for first series

Answer 21

So I also just realized I had the derivatives written different, I had \[y''= \sum_{0}^{\infty} (n-1)na _{n}(x-3)^{n-2}\], is this wrong?

Answer 22

Hey wait, there is a mistake in my derivatives

Answer 23

Haven't you noticed my mistake yet ?

Answer 24

I think I noticed it above, right?

Answer 25

see the mistake ?

Answer 26

You're missing the n?

Answer 27

Yes

Answer 28

Lets fix the differential equation quick

Answer 29

Are there two different ways to do the derivatives? Because I wrote mine differently before

Answer 30

Here is the corrected equation :
\[\sum\limits_{n=2}^{\infty}a_n n(n-1)(x-3)^{n-2}\\
- \sum\limits_{n=1}^{\infty}a_n n(n-1)(x-3)^{n+1} -6 \sum\limits_{n=1}^{\infty}a_n n(n-1)(x-3)^{n}-17 \sum\limits_{n=1}^{\infty}a_n n(n-1)(x-3)^{n-1}\\
+ \sum\limits_{n=1}^{\infty}a_n(x-3)^{n}\\
=0\]

Answer 31

Okay that makes sense

Answer 32

Nope, there is only one way to do differentiation.
See if above looks good

Answer 33

Now try making all the exponents equal to (x-3)^n

Answer 34

So I would change the lower bound for the first term to negative two to get the exponent the same?

Answer 35

I'll show you how to fix the exponent for the first series

Answer 36

\(\sum\limits_{n=2}^{\infty}a_n n(n-1)(x-3)^{n-2}\)

Let \(k=n-2\)
this gives \(n = k+2\) and the lower bound becomes \(k=2-2=0\)

the series is then :
\(\sum\limits_{k=0}^{\infty}a_{k+2} (k+2)(k+1)(x-3)^{k}\)

Answer 37

Btw, index variable is just dummy, it doesn't matter what variable you use for index. So you can replace k by n in the end

Answer 38

\(\sum\limits_{k=0}^{\infty}a_{k+2} (k+2)(k+1)(x-3)^{k}\)

is same as

\(\sum\limits_{n=0}^{\infty}a_{n+2} (n+2)(n+1)(x-3)^{n}\)

Answer 39

try fixing the exponents of remaining series

Answer 40

okay gotcha

Answer 41

So would the next part be \[\sum_{2}^{\infty} a_{n-1}(n-1)(n-1)(x-3)^n\]

Answer 42

wait, but the second (n-1) should be (n-2) my bad

Answer 43

Lets see

Answer 44

\(\sum\limits_{n=1}^{\infty}a_n n(n-1)(x-3)^{n+1}\)

Let \(n+1=k\)
this gives \(n=k-1\) and the lower bound becomes \(k=1+1=2\)

the series is then
\(\sum\limits_{k=2}^{\infty}a_{k-1} (k-1)(k-2)(x-3)^{k}\)

Answer 45

that is same as
\(\sum\limits_{n=2}^{\infty}a_{n-1} (n-1)(n-2)(x-3)^{n}\)

Answer 46

okay awesome, I think I got all of them then. So now I need to change the lower bound to n=2 because that's the highest one, correct?

Answer 47

Exactly!

Answer 48

which is just pulling out factors

Answer 49

To do that, just observe this split up :
\[\sum\limits_{n=0}^{\infty}a_n x^n = a_0+a_1x + \sum\limits_{n=2}^{\infty}a_n x^n \]

Answer 50

Yes its just pulling out first few terms

Answer 51

okay sweet, ill try the first one then

Answer 52

Okay so I get \[2a_{2} + 6a_{3}(x-3) + \sum_{n=2}^{\infty} a_{n+2}(n+2)(n+1)(x-3)^n\] for the first term

Answer 53

And then for the second term I don't do anything, and for the third term plugging in 1 for n will give me zero?

Answer 54

Looks perfect. Yes, for the third term :
\[\sum\limits_{n=1}^{\infty}a_n n(n-1)(x-3)^{n}= 0+ \sum\limits_{n=2}^{\infty}a_n n(n-1)(x-3)^{n}\]

Answer 55

Hey sorry wait

Answer 56

So the bottom part is what I have right now, is that correct?

Answer 57

scratch all that, there is a mistake in our derivative for y' itself

Answer 58

oh but the x^n should be (x-3)^n at the end

Answer 59

ohhhhh

Answer 60

Wrong equation :
\[\sum\limits_{n=2}^{\infty}a_n n(n-1)(x-3)^{n-2}\\
- \sum\limits_{n=1}^{\infty}a_n n\color{red}{(n-1)}(x-3)^{n+1} -6 \sum\limits_{n=1}^{\infty}a_n n\color{red}{(n-1)}(x-3)^{n}-17 \sum\limits_{n=1}^{\infty}a_n n\color{red}{(n-1)}(x-3)^{n-1}\\
+ \sum\limits_{n=1}^{\infty}a_n(x-3)^{n}\\
=0\]

Correct equation :
\[\sum\limits_{n=2}^{\infty}a_n n(n-1)(x-3)^{n-2}\\
- \sum\limits_{n=1}^{\infty}a_n n(x-3)^{n+1} -6 \sum\limits_{n=1}^{\infty}a_n n(x-3)^{n}-17 \sum\limits_{n=1}^{\infty}a_n n(x-3)^{n-1}\\
+ \sum\limits_{n=1}^{\infty}a_n(x-3)^{n}\\
=0\]

Answer 61

you will need to work from above equation again, sry about that :)

Answer 62

no problem. so from there just change the exponents again?

Answer 63

Yes

Answer 64

1) change the exponent of all series to (x-3)^n
2) have same lower bound for all series by pulling out first few terms

Answer 65

okay ill do that now

Answer 66

Found another mistake

Answer 67

Here is the final eqn with all mistakes corrected, hopefully :
\[
\sum\limits_{n=2}^{\infty}a_n n(n-1)(x-3)^{n-2}\\
- \sum\limits_{n=1}^{\infty}a_n n(x-3)^{n+1} -6 \sum\limits_{n=1}^{\infty}a_n n(x-3)^{n}-17 \sum\limits_{n=1}^{\infty}a_n n(x-3)^{n-1}\\
+ \sum\limits_{n=\color{red}{0}}^{\infty}a_n(x-3)^{n}\\
=0\]

Answer 68

alright sounds good, I hadn't reached there yet anyways haha

Answer 69

\[2a_{2}+a_{0}+6a_{3}(x-3)-5a_{1}(x-3)+\sum_{n=2}^{\infty}[a_{n+2}(n+2)(n+1)-a_{n-1}(n-1)-6a_{n}n-17a_{n+1}(n+1)+a_{n}](x-3)^n\]

This is what I got from that

Answer 70

Looks you have forgot the pulled out terms from fourth series :
\[-17a_1-34a_2(x-3)\]

Answer 71

oh right I wrote it down just forgot it in the equation writer haha

Answer 72

This problem is so lenghty, but I see you have the right ideas... I hope they don't give you this in exam

Answer 73

so now I can set the entire inside of that sum equal to zero and that will be my RR for n=2,3,...

Answer 74

I dont think they will, i hope not

Answer 75

Yes, before that set the coeffecients of (x-3)^0, (x-3)^1 equal to 0 and solve whatever coefficents you can

Answer 76

and then I can set a0=c1 and a1=c2 because it's a second order equation?

Answer 77

Yes

Answer 78

setting the coefficient of (x-3)^0 equal to 0 :
\(2a_2+a_0-17a_1 = 0\)

\(\implies a_2 = \dfrac{-c_1+17c_2}{2}\)

Answer 79

setting the coefficient of (x-3)^1 equal to 0 :
\(6a_3-5a_1-34a_2 = 0\)
\(\implies a_3=\cdots \)

Answer 80

okay sweet, and I just need the first 3 terms of a for the answer

Answer 81

\[a_{3}=[5c_{2}-17(17c_{2}-c_{1})]/6\]
would be the third term then?

Answer 82

it says I need the first 3 terms in each of two linearly independent solutions

Answer 83

For the first solution, set \(c_1 = 1, ~c_2=0\)

For the second solution, set \(c_1 = 0, ~c_2=1\)

Answer 84

so just plug in those values and my answer will be the results?