Question

Why are there x's in the numerator?

Answer 1

why not? lol

Answer 2

it looks like a taylor or maclurian series it just has x.

Answer 3

http://en.wikipedia.org/wiki/Taylor_series

it is just the pattern

Answer 4

Looks like a power series solution to a differential equation.
Learning these now? :D

Oh boy those were a pain, lot of steps.. generating that list of terms...

Answer 5

Here is how the problem started out....well not really, but this is how far I got.

\[n=0,a_2=\frac{a_0}{2}\]
\[n=1,a_3=\frac{a_1}{3}\]
\[n=2,a_4=\frac{a_0}{2\cdot4}\]
\[n=3,a_5=\frac{a_1}{3\cdot5}\]
\[n=4,a_6=\frac{a_0}{2\cdot4\cot6}\]
\[n=5,a_7=\frac{a_1}{3\cdot5\cdot7}\]
\[n=6,a_8=\frac{a_0}{2\cdot4\cdot6\cdot8}\]
\[n=7,a_9=\frac{a_1}{3\cdot5\cdot7\cdot9}\]
\[n=8,a_10=\frac{a_0}{2\cdot4\cdot6\cdot8\cdot10}\]

Answer 6

so why do we magically make x's appear in the numerator?

Answer 7

I'm guessing it came from the given ODE?

Answer 8

\[y''-xy'-y=0\]

Answer 9

\[\huge y=\sum_{i=0}^\infty C_n \cdot x^n\]Because your solution will be of this form Jennifer.

When you expand this out, and then plug in all your C values, that's where the X's are coming from.

Answer 10

interesting... so why didn't I have to do that for y''-y=0?

Answer 11

I did that problem with @zarkon http://openstudy.com/users/jennifersmart1#/updates/50cfbed4e4b06d78e86d61e4

Answer 12

Hmm yah that one doesn't appear to be correct :(
The x's are missing.

At the end of your opening message you say,
"I understand that I will have \[y=a_0\sum +a_1\sum\]

There's nothing written in those sums D:
You're solution should be a function of x.

Answer 13

I see :)

Answer 14

\[\huge y=\sum_{n=0}^{\infty}a_n \cdot x^n\]Maybe this notation will make more sense to you, with the a's instead of c's.

Answer 15

After you plug in all of your A terms, you'll have the entire series in terms of A0 and A1, from there you'll be able to write it as TWO sums the way you originally thought.

Answer 16

Ok makes sense, let's see here...
give me a min...I'll post the solution I come up with

Answer 17

ok I'm still a little confused.
I'm able to get this part....so now how do I figure out what exponents the x's have?
\[a_0\left(\frac{1}{2}+\frac{1}{2\cdot4}+\frac{1}{2\cdot4\cdot6}+\frac{1}{2\cdot4\cdot6\cdot8}+\frac{1}{2\cdot4\cdot6\cdot8\cdot10}\right)\]

Answer 18

\[\large y=\sum_{n=0}^{\infty}a_n \cdot x^n\]\[\large =a_0x^0+a_1x^1+a_22x^2+a_3x^3+...+a_nx^n\]
Your series solution should be of this form.
If you simply expand out this sum, you'll see that the A values match the X values.
I'm not sure how you're able to split up the A0's and A1's without expanding out this solution first :O hmm

Answer 19

I guess that's a better way to start =D
Ok I'll try it again

Answer 20

\[\large = a_0+a_1x+\left(\frac{a_0}{2}\right)x^2+\left(\frac{a_1}{3}\right)x^3+...\]See how I'm plugging in the A values? :o
You'll be able to write this entire series solution in terms of A0 and A1 using the list you generated.
Split it up after that :)

At least that's the way I learned to do it heh.

Answer 21

oh ok. I will

Answer 22

Hmm I'm getting the same A values when I generate a list of terms from the original problem.

So you're probably on the right track! :D

Answer 23

almost done.... :D

Answer 24

I'm just a little sloooowwww

Answer 25

Hah XD Can't fault a Koala for moving slow...

Answer 26

LOL!!!! that was AWESOME hahaha

Answer 27

brb :3

Answer 28

help plz =C

\[a_0+a_1x^1+a_0x^2+a_1x^3+\frac{a_0}{2}x^4+\frac{a_1}{3}x^5+\frac{a_0}{2\cdot4}x^6+\frac{a_1}{3\cdot5}x^7\]+\[\frac{a_0}{2\cdot4\cdot6}x^8+\frac{a_1}{3\cdot5\cdot7}x^9+\frac{a_0}{2\cdot4\cdot6\cdot8}x^{10}\]

Answer 29

I think the A2 term was A0/2

It looks like you have everything shifted over one too far.\[a_0+a_1x^1+\frac{a_0}{2}x^2+\frac{a_1}{3}x^3+\frac{a_0}{2\cdot4}x^4+\frac{a_1}{3\cdot5}x^5+\frac{a_0}{2\cdot4\cdot6}\]

Answer 30

Right? :O
You just did something silly when you plugged everything in.
You plugged in the A2 value for the A4, and so on.

Answer 31

I disagree

Answer 32

because it's \[a_22x^2\] and the those cancel out
\[\frac{a_0}{2}2x^2\]

Answer 33

Where did that extra 2 come from? :o

Answer 34

\[\large =a_0x^0+a_1x^1+a_22x^2+a_3x^3+...+a_nx^n\]\
that's what you wrote LOL

Answer 35

Oh, just a typo :\

The series solution should only be generating A's and X's :)

Answer 36

My bad :c

Answer 37

oh ok :)
I'll write that again...correctly this time ;P

Answer 38

Don't just believe everything I say!! Make sure it makes sense XD lol

Answer 39

\[a_0+a_1x^1+a_0x^2+a_1x^3+\frac{a_0}{2\cdot4}x^4+\frac{a_1}{3\cdot5}x^5+\frac{a_0}{2\cdot4\cdot6}x^6+\frac{a_1}{3\cdot5\cdot7}x^7\]+\[\frac{a_0}{2\cdot4\cdot6\cdot8}x^8+\frac{a_1}{3\cdot5\cdot7\cdot9}x^9+\frac{a_0}{2\cdot4\cdot6\cdot8\cdot10}x^{10}\]

Answer 40

oops i forgot some terms

Answer 41

\[a_0+a_1x^1+\frac{a_0}{2}x^2+\frac{a_1}{3}x^3+\frac{a_0}{2\cdot4}x^4+\frac{a_1}{3\cdot5}x^5+\frac{a_0}{2\cdot4\cdot6}x^6+\frac{a_1}{3\cdot5\cdot7}x^7\]+
\[\frac{a_0}{2\cdot4\cdot6\cdot8}x^8+\frac{a_1}{3\cdot5\cdot7\cdot9}x^9+\frac{a_0}{2\cdot4\cdot6\cdot8\cdot10}x^{10}\]

Answer 42

\[\huge \color{gray}{C(} \;^o O\color{green}{\text{}}\;^o\color{gray}{) D} \qquad \large \text{Wait a minute...}\]

Answer 43

Ah nevermind, yes that looks good :)

Answer 44

Is that a Koala?

Answer 45

I was trying yah :C hard to draw koala in alphabet lol

Answer 46

lol nice work!

Answer 47

So from here, we can separate the A0 and A1 terms.

Answer 48

sure
\[a_0+\frac{a_0}{2}x^2++\frac{a_0}{2\cdot4}x^4+\frac{a_0}{2\cdot4\cdot6}x^6+\frac{a_0}{2\cdot4\cdot6\cdot8}x^8+\frac{a_0}{2\cdot4\cdot6\cdot8\cdot10}x^{10}\]
+
\[a_1x^1+\frac{a_1}{3}x^3+\frac{a_1}{3\cdot5}x^5+\frac{a_1}{3\cdot5\cdot7}x^7+\frac{a_1}{3\cdot5\cdot7\cdot9}x^9\]

Answer 49

Good good. If we factor out the A constants, we can start to build our Summation.
\[a_0\left(1+\frac{x^2}{2}+\frac{x^4}{2\cdot4}+\frac{x^6}{2\cdot4\cdot6}+...\right)\]

Answer 50

I think I see a pattern. OK soo ....

Answer 51

hmmm

Answer 52

Hmm I like this problem.. the numbers will work out really nicely.

Answer 53

\[\huge a_0\left(\sum_{n=0}^\infty\frac{x^{\color{orangered}{n}}}{\color{orangered}{?}}\right)\]Hmm so we need to fix these red parts.
The powers are even on the A0 term, maybe we can do something about the exponent on the x.

Answer 54

Here is what is going wrong with my brain....so I see \[\frac{x^4}{2\cdot4}\]
and I think n=4 here and 2x4 is the ?
...that doesn't make sense...sorry I don't mean to be soo difficult =(

Answer 55

Yah I'm having trouble with that part also.
Our teacher let us be lazy and just write it as, 2*4*6*...

The solution step you posted earlier, was this for this problem or just something else?

Answer 56

Cuz they did something fancy, and I'm trying to figure it out :D

Answer 57

it's from chegg.com....someone posted that as the solution

Answer 58

oh heh

Answer 59

he was given 4 stars for his solution

Answer 60

Does the top make sense though?
In the A0 series, we only want to generate EVEN powers on X.
So if we write it as,\[\huge a_0\left(\sum_{n=0}^\infty\frac{x^{2n}}{\color{orangered}{?}}\right)\]
That should give us only even powers of x, right?

Answer 61

yeah that makes sense

Answer 62

For the bottom I guess what we have is this,\[\large 2\cdot4\cdot6\cdot8\cdot \cdot \cdot (2n)\quad = \quad 2(1\cdot2\cdot3\cdot4\cdot \cdot \cdot (n))\]

Answer 63

It looks like what we're generating on the bottom is,\[\large 2(n)!\]I'm not sure where they got 2^n from. Lemme think, hmm

Answer 64

Yah that shouldn't be 2^n. If you expand out just a few terms you can see that it's way off. I think I have it right... hopefully.

Answer 65

check this out!!!. here is the solution in the back of the book

Answer 66

Are we on number 9?

Answer 67

number 9

Answer 68

Ugh m brain failing me :C

They have it written as a single power series. Did they do some super simplification? Or did we mess up our A values? hmm

Answer 69

Hmmm let's see...I can write out everything....you can check it. It will take me a min to write out

Answer 70

\[y''-xy'-y=0\]
\[y=\sum_{n=0}^{\infty}a_nx^n\]
\[y'=\sum_{n=1}^{\infty}a_nnx^{n-1}\]
ok this is gonna take me a while...

Answer 71

lolol yah it might not be worth your time XD

Answer 72

for one stupid problem :D heh

Answer 73

The last approach is the one I would of taken. Performed shifts to get n = 0
and so forth.

Answer 74

@abb0t explain

Answer 75

I got \[a_{n+2}=\frac{a_n}{n+2}\] and I wrote it out in a previous post...could that be where I went wrong?

Answer 76

Used the formula for y,
found y' and y''
substituted in y, y' and y'' into your equation. shifted n = 2 for y'' and n = 1 for y' to n = 0 so that you can combine everything and solve for y.

Answer 77

\[\large a_{n+2}=\frac{a_n}{n+2} \; \text{for }n \ge1\]
After shifting and combining, I came up with the same jenn.. so i dunno..

Answer 78

\[y '' = \sum_{n=2}^{∞} n(n-1)a_n x^{n-2}\]
shift:
\[\sum_{n=0}^{∞}(n+2)(n+1)a_{n+2}x^n\]

Answer 79

that's what I got too
\[\sum_{n=0}^{∞}(n+2)(n+1)a_{n+2}x^n-na_nx^n-a_nx^n\]

Answer 80

=0
of course

Answer 81

i'm about to get kicked out of my study place....see you all in an hour or soo....
THanks everyone for all of your help!!!! hope to see you soon =D

Answer 82

There's an x in front of y'.

Answer 83

\[\large x \sum_{n=1}^\infty n a_n x^{n-1}= \sum_{n=1}^\infty n a_n x^n\]Yah I think we through that part alright. hmm

Answer 84

i'm back

Answer 85

\[y=\sum_{n=0}^{\infty} a_nx^n\]
\[y'=\sum_{n=1}^{\infty} a_nnx^{n-1}\]
\[xy'=\sum_{n=1}^{\infty}na_nx^n\]
\[y''=\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}\]
\[\sum_{n=0}^{\infty}\left[(n+2)(n+1)a_{n+2}-na_n-a_n\right]x^n=0\]

Answer 86

@AccessDenied

Answer 87

@TuringTest

Answer 88

Hey Turing, can you please take a look at this in the morning? I did everything right up until the point where I had to write it in Sum form. The solution in the back of my book (I attached a screen shot) didn't split the sum up in two. Which makes it even more confusing.

Answer 89

@UnkleRhaukus

Answer 90

where are we

Answer 91

ok,
\[y=\sum_{n=0}^{\infty} a_nx^n\]
\[y'=\sum_{n=1}^{\infty} a_nnx^{n-1}\]
\[xy'=\sum_{n=1}^{\infty}na_nx^n\]
\[y''=\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}\]
\[y''-xy'-y=0\]
\[\sum_{n=0}^{\infty}\left[(n+2)(n+1)a_{n+2}-na_n-a_n\right]x^n=0\]
\[a_{n+2}=\frac{a_n}{n+2}\]
\[n=0,a_2=\frac{a_0}{2}\]
\[n=1,a_3=\frac{a_1}{3}\]
\[n=2,a_4=\frac{a_0}{2\cdot4}\]
\[n=3,a_5=\frac{a_1}{3\cdot5}\]
\[n=4,a_6=\frac{a_0}{2\cdot4\cdot6}\]
\[n=5,a_7=\frac{a_1}{3\cdot5\cdot7}\]
\[n=6,a_8=\frac{a_0}{2\cdot4\cdot6\cdot8}\]
\[n=7,a_9=\frac{a_1}{3\cdot5\cdot7\cdot9}\]
\[n=8,a_{10}=\frac{a_0}{2\cdot4\cdot6\cdot8\cdot10}\]
\[y=a_0+\frac{a_0}{2}x^2++\frac{a_0}{2\cdot4}x^4+\frac{a_0}{2\cdot4\cdot6}x^6+\frac{a_0}{2\cdot4\cdot6\cdot8}x^8+\frac{a_0}{2\cdot4\cdot6\cdot8\cdot10}x^{10}\]
+\[a_1x^1+\frac{a_1}{3}x^3+\frac{a_1}{3\cdot5}x^5+\frac{a_1}{3\cdot5\cdot7}x^7+\frac{a_1}{3\cdot5\cdot7\cdot9}x^9\]
\[a_0\left(1+\frac{x^2}{2}+\frac{x^4}{2\cdot4}+\frac{x^6}{2\cdot4\cdot6}+...\right)\]

Answer 92

how do I write the sum?

Answer 93

you just want the last line converted into summation notation?

Answer 94

forgot to type \[y=a_0\left(1+\frac{x^2}{2}+\frac{x^4}{2\cdot4}+\frac{x^6}{2\cdot4\cdot6}+...\right)\]
+ the a_1 which have the odd numbers in the denominator
but how do I go about doing that?