Question

how did they determine

Answer 1

https://gyazo.com/32b7234d6e962e0ad9be42a08972b285

Answer 2

b_0
b_1
b_2
b_k

Answer 3

HI
you are thinking tooo hard

Answer 4

any polynomial can be written as a power series, most of the coefficients are zero

Answer 5

If you want to be more systematic, you can plug that power series representation in to the differential equation and then match coefficients.

Answer 6

@dan815

Answer 7

i would follow what @Jemurray3 said

Answer 8

watchu u mean its just what they say

Answer 9

1+2(x-1) + 1*(x-1)^2 + 0*(x-1)^3 + 0*(x-1)^4....
so the power series
sum bk*(x-1)^k
where
b0=1
b1=2
b2=1
bk, k>2 is all 0

Answer 10

i mean how did they get it

Answer 11

solution pls!

Answer 12

like do i plug in the whole power series to get it

Answer 13

this is what i get:
x = 0;
b_0(0-1)^0 = 2
b_0 1 = 2
b_ 0 = 2 ..

Answer 14

http://prntscr.com/8o6l00

Answer 15

they just found a simple solution with (x-1) power series

Answer 16

so they didnt bother with the rest

Answer 17

but normally if u cant see that u would plug in both of the power series and solve for the coefficients

Answer 18

you know the summation has to equal x^2, so just math the coefficients

Answer 19

Here's how you do it. If \( y = \sum_{k=0}^\infty b_k (x-1)^k \), then
\[ y' = \sum_{k=1}^\infty k\cdot b_k (x-1)^{k-1} \]
Notice how I replaced the bottom limit with 1, because the 0 term is just 0 now.

Next, you rename the index by letting \( n = k-1 \rightarrow k = n+1 \). That means
\[y' = \sum_{n=0}^\infty (n+1)\cdot b_{n+1} (x-1)^n \]

Now, the left hand side of the differential equation is \(y+y'\). So we can combine those two terms together using the same index:

\[y + y' = \sum_{k=0}^\infty b_k (x-1)^k + \sum_{k=0}^\infty (k+1)\cdot b_{k+1} (x-1)^k\]
\[ = \sum_{k=0}^\infty \left\{ b_k + (k+1)b_{k+1}\right\}(x-1)^k \]
The first few terms if you expand that out are
\[(b_0 + b_1) +(b_1 + 2b_2)(x-1) + (b_2 + 3b_3)(x-1)^2 + ... \]
At the same time, you have the right hand side:
\[1 + 2(x-1) + (x-1)^2 \]