Question

solve by using power series method: 2y'-y=cosh(x). compute the first 6 coefficients (a0-a5)

Answer 1

Okay, that's a little long.
What part of it don't you get?

Answer 2

yeah, i have doubt in one part... I'll show you

Answer 3

Assume that y has power series

y= ∑_(n=0)^∞▒〖a_n x^n 〗
Then
y^'= ∑_(n=0)^∞▒〖na_n x^(n-1) 〗
so that
2y^'-y=∑_(n=0)^∞▒〖2na_n x^(n-1)-a_n x^n 〗

Collect the above series with powers of x.

∑_(n=0)^(∞ )▒〖 █(2na_n x^(n-1)-a_n x^n=2(a_1+2a_2 x+3a_3 x^2+4a_4 x^3+5a_5 x^4…)-@(a_0+a_1 x+a_2 x^2+a_3 x^3+a_4 x^4+a_5 x^5…)@)〗


=(2a_1-a_0 )+(2∙2a_2-a_1 )x+(2∙3a_3-a_2 ) x^2+(2∙4a_4-a_3 ) x^3+⋯

After that, I'm confused I don't know if is:
cosh⁡x=1+x^2/2!+x^4/4!+x^6/6!+⋯ 〗
or cosh⁡x=1+x^2/2+x^4/6+x^6/24+x^8/120⋯ 〗
I think im having problems with the algebra

Answer 4

I'll just give you the final answer.

Answer 5

..........because I can't see the stuff you typed up......

Answer 6

Have you seen my other post on the almost identical problem?

Answer 7

with the senh(x)?

Answer 8

http://openstudy.com/updates/51c4e47de4b069eb00c54770

Answer 9

Do you know the power series for \(\cos x\)?$$\cos x=\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}x^{2n}$$