how did they determine

- Mimi_x3

how did they determine

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- chestercat

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- Mimi_x3

https://gyazo.com/32b7234d6e962e0ad9be42a08972b285

- Mimi_x3

b_0
b_1
b_2
b_k

- misty1212

HI
you are thinking tooo hard

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## More answers

- misty1212

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

- anonymous

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

- Mimi_x3

@dan815

- anonymous

i would follow what @Jemurray3 said

- dan815

watchu u mean its just what they say

- dan815

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

- Mimi_x3

i mean how did they get it

- Mimi_x3

solution pls!

- Mimi_x3

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

- Mimi_x3

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

- dan815

http://prntscr.com/8o6l00

- dan815

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

- dan815

so they didnt bother with the rest

- dan815

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

- dan815

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

- anonymous

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:
\[1 + 2(x-1) + (x-1)^2 \]

- dan815

|dw:1444088025451:dw|

- dan815

does it make sense now?

- dan815

u can get expreesion for ak

- anonymous

Oops - I think I misread the intent of the question.

- dan815

|dw:1444088417372:dw|

- dan815

and the rest of the qeuations equal 0

- dan815

at least u will get 3 equations and 4 unkowns
so you will get some constant of variation but all your Ak can be written as a function of 1 variable

- dan815

solve for a0,a1,a2,a3

- dan815

|dw:1444102770646:dw|

- Mimi_x3

I'm finding B_0

- dan815

what do u mean finding B_0 that is given

- dan815

do u mean u are finding Ao

- dan815

|dw:1444102878452:dw|

- Mimi_x3

b_0 is given frmo the solution...but i don't how they got it.

- Mimi_x3

THIS IS TH QUESTION https://gyazo.com/e098e1ac11c87668eda64d1a3a316ee2

- dan815

|dw:1444103319383:dw|

- dan815

|dw:1444103402756:dw|

- dan815

@Empty

- dan815

@oldrin.bataku

- anonymous

$$x^2=(x-1)^2+2(x-1)+1$$ we also know $$x^2=\sum_{n=0}^\infty b_n (x-1)^n$$ so it follows \(b_0=1,b_1=2,b_2=1\) and \(b_k=0\) for \(k\ge 3\)

- anonymous

you can determine that by many different means, including the Taylor coefficient formula: $$b_n=\frac{f^{(n)}(1)}{n!}$$ so $$f(x)=x^2\implies f(1)=1\\f'(x)=2x\implies f'(1)=2\\f''(x)=2\implies f''(1)=2\\f^{(3)}(x)=0\implies f^{(3)}(1)=0\\f^{(n)}(x)=0\implies f^{(n)}(1)=0\text{ for }n\ge 3$$

- anonymous

or just by expanding, like I did above: $$(x-1)^2=x^2-2x+1\\\implies x^2=(x-1)^2+2x-1\\2(x-1)=2x-2\\\implies x^2=(x-1)^2+2(x-1)+1$$

- Mimi_x3

ok i got it
but i dont' understand how they determine boundaries it feels like magic
https://gyazo.com/848f27d09ebe1758077d624b8cc6ab0e
how
y(0) = 0 = a_0 (all even coef)

- Mimi_x3

and y'(0) = 1 = a_1

- anonymous

in that problem, consider: $$y(x)=\sum_{n=0}^\infty a_n x^n\\y(x)=a_0+a_1x+a_2x^2+\dots\\y(0)=a_0+a_1\cdot 0+a_2\cdot 0+\dots=a_0$$ so if \(y(0)=0\) is given as the BC, then it follows \(a_0=0\)
similarly, $$y(x)=\sum_{n=0}^\infty a_n x^n\\y'(x)=\sum_{n=0}^\infty na_n x^n\\y'(x)=a_1+2a_2 x+3a_3 x^2+\dots\\y'(0)=a_1+2a_2\cdot 0+3a_3\cdot 0+\dots=a_1$$so \(y'(0)=1\) gives us \(a_1=1\)

- anonymous

setting \(x=0\) makes every term in the power series with nonzero power equal \(0\), and for convenience here we define \(x^0=1\) even at \(x=0\)

- Mimi_x3

thanks! you're awesome!!

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