Question

\[a_{k+2}=a_{k+1}\frac{k+1}{k+2}\]
\[a_{1}\frac{1}{2}
+a_{2}\frac{2}{3}
+a_{3}\frac{3}{4}
+a_{4}\frac{4}{5}
+a_{5}\frac{5}{6}
+a_{6}\frac{6}{7}
+a_{7}\frac{7}{8}
+a_{8}\frac{8}{9}
+a_{9}\frac{9}{10}
+a_{10}\frac{10}{11}
+a_{11}\frac{11}{12}\]

Answer 1

I'm not sure where you're going with the sum, you're jumping ahead too much and losing track of where we were :)

Answer 2

oh and we said \[a_2=\frac{a_1}{2}\]

Answer 3

Get your list straightened out first, everything in terms of A1.

Answer 4

\[\frac{a_1}{2}
+\frac{a_1}{3}+\dots\]

Answer 5

I forgot how to do a_3

Answer 6

\[k=1 \quad \rightarrow a_3=2a_2/3 \quad \rightarrow \quad a_3=a_1/3\]\[k=2 \quad \rightarrow a_4=3a_3/4 \quad \rightarrow \quad a_4=a_1/4\]\[k=3 \quad \rightarrow a_5=4a_4/5 \quad \rightarrow \quad a_5=a_1/5\]

Answer 7

This is the list that we're generating.

Answer 8

Confused how I got from the middle step to the last?

Answer 9

Wow...I did get a little ahead of myself....
No it makes perfect sense

Answer 10

You're forgetting all the X values in your series solution. Remember when we went over that before? :)

Our solution should be of the form,\[\large \sum_{n=0}^\infty a_n x^n\]So when you expand that out, you should be getting powers of X along with your A's

Answer 11

\[k=1 \quad \rightarrow a_3=2a_2/3 \quad \rightarrow \quad a_3=a_1/3\]
\[k=2 \quad \rightarrow a_4=3a_3/4 \quad \rightarrow \quad a_4=a_1/4\]
\[k=3 \quad \rightarrow a_5=4a_4/5 \quad \rightarrow \quad a_5=a_1/5\]
what does \quad do?

Answer 12

it makes a nice big space. \qquad makes a larger space. \; makes a small space.

Answer 13

I see

Answer 14

Oh actually, since we started with y' as lowest degree of y, i think our solution will start at n=1....

Otherwise we end up with an A0 term that we can't use.

Answer 15

\[k=1 \quad \rightarrow a_3=2\frac{a_2}{3} \quad \rightarrow \quad a_3=\frac{a_1}{3}\]
\[k=2 \quad \rightarrow a_4=3\frac{a_3}{4} \quad \rightarrow \quad a_4=\frac{a_1}{4}\]
\[k=3 \quad \rightarrow a_5=4\frac{a_4}{5} \quad \rightarrow \quad a_5=\frac{a_1}{5}\]
\[k=4 \quad \rightarrow a_6=5\frac{a_5}{6} \quad \rightarrow \quad a_6=\frac{a_1}{6}\]
\[k=5 \quad \rightarrow a_7=6\frac{a_6}{7} \quad \rightarrow \quad a_7=\frac{a_1}{7}\]
\[k=6 \quad \rightarrow a_8=7\frac{a_7}{8} \quad \rightarrow \quad a_8=\frac{a_1}{8}\]
\[k=7 \quad \rightarrow a_9=8\frac{a_8}{9} \quad \rightarrow \quad a_9=\frac{a_1}{9}\]
\[k=8 \quad \rightarrow a_{10}=9\frac{a_9}{10} \quad \rightarrow \quad a_{10}=\frac{a_1}{10}\]
\[k=9 \quad \rightarrow a_{11}=10\frac{a_{11}}{12} \quad \rightarrow \quad a_{11}=\frac{a_1}{11}\]
\[k=10 \quad \rightarrow a_{12}=12\frac{a_{12}}{13} \quad \rightarrow \quad a_{12}=\frac{a_1}{12}\]

Answer 16

Yah I think you have the right pattern! That's what I was coming up with at least.

Answer 17

\[y\quad =\quad \sum_{n=1}^\infty a_n x^n \quad =\]\[a_1x^1+a_2x^2+a_3x^3+a_4x^4+...\]

Answer 18

Now just plug in your a values!

Answer 19

\[a_3=\frac{a_1}{3}+\quad
a_4=\frac{a_1}{4}+\quad
a_5=\frac{a_1}{5}+\quad
a_6=\frac{a_1}{6}+\quad
a_7=\frac{a_1}{7}+\quad
a_8=\frac{a_1}{8}+\quad
a_9=\frac{a_1}{9}+\quad
a_{10}=\frac{a_1}{10}+\quad
a_{11}=\frac{a_1}{11}+\quad
a_{12}\frac{a_1}{12}\]
I'm sorry I'm really tired....

brain freeze...what do you mean by 'plug in your a values'

Answer 20

What I posted a moment ago, is the FORM your solution will be in.

\[a_1x^1+a_2x^2+a_3x^3+a_4x^4+...\]Plugging in the value for a_2 gives us,\[a_1x^1+\frac{a_1}{2}x^2+a_3x^3+a_4x^4+...\]

Answer 21

sry :( rly tired

oh my....
ok, so.... what is our goal?....I lost sight of it again....
Oh yeah...
we're trying to write a sum describing this above pattern...

We'll reach that goal by....

Answer 22

I know I have some grammatical errors too...LOL

Answer 23

lol :D poor gal

Answer 24

\[y\quad =\quad \sum_{n=1}^\infty a_n x^n \quad =\]
\[a_1x^1+a_2x^2+a_3x^3+a_4x^4+...\]
ohhhh
I plug in the a values!!!! haha

Answer 25

Remember back to those friendly linear homogeneous DE's, we knew the answer would be of the form \[\large y=c_1 e^{r_1x}+c_2e^{r_2x}\]Something like that...

Well with power series, we know that our solution will be of this form,\[y\quad =\quad \sum_{n=1}^\infty a_n x^n\]
We expand out this power series, and from there we want to rewrite it as a series with ONLY 1 A value, not a bunch of different A values.

Answer 26

\[y\quad =\quad \sum_{n=1}^\infty a_n x^n \quad =\]
\[a_1x^1+\frac{a_1}{2}x^2+\frac{a_1}{3}x^3+\frac{a_1}{4}x^4+\frac{a_1}{5}x^5+\frac{a_1}{6}x^6+\frac{a_1}{7}x^7+\frac{a_1}{8}x^8+\frac{a_1}{9}x^9+\frac{a_1}{10}x^{10}\]

Answer 27

Yah looks good c: So every term is written in terms of A1.

So now we can rewrite it as a nice clean summation.

Answer 28

Factoring out the A1 before we do that though*

Answer 29

\[a_1\left(x^1+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+\frac{x^6}{6}+\frac{x^7}{7}+\frac{x^8}{8}+\frac{x^9}{9}+\frac{x^{10}}{10}\right)\]

Answer 30

Notice a pattern? :) It's a nice simple one this time.

Answer 31

\[\huge =a_1 \sum_{n=1}^\infty \frac{x^?}{?}\]

Answer 32

\[y\quad =\quad a_1\sum_{n=1}^\infty \frac{x^n}{n} \quad \]

Answer 33

Yayyyyy \:D/

Answer 34

Hopefully we didn't make any mistakes along the way :U You never know!

Answer 35

we forgot a negative sign according to chegg.com

Answer 36

nope...sorry....my fault...we're right

Answer 37

We are? :o oh yay.

Answer 38

Oh ok, so we ARE suppose to start at 0, and then we just leave the A0 alone in front. I guess that makes sense :)

Answer 39

I'll revisit that in the morning

thanks for all of your help!!!

Answer 40

np \c:/ haf a good nite

Answer 41

and Merry Christmas! :D

Answer 42

\[\huge \text{YOU} \quad \text{TOO}\quad =D\]