Question

need to solve y' -y=x by power series method

Answer 1

if I remember correctly (please check) you need to present y as:
\[y=\sum_{0}^{\infty}c _{n}x ^{n}\]
y in the form of power series in x

Answer 2

That's correct.

Answer 3

then y':
\[y'=\sum_{0}^{\infty}nc _{n}x ^{n-1}=\sum_{1}^{\infty}c _{n}x ^{n-1}\]
can you pick up from here?

Answer 4

\[c_1 = c_0\]
\[c_n = \frac{1+c_0}{n!}\]
anyone second this?

Answer 5

i got :
\[c _{n}= c _{0}/n!\]
...?

Answer 6

looking at the first few terms
\[c_1 - c_0 = 0\]
so
\[c_1 = c_0.\]
Then
\[2c_2 - c_1 = 1\]
so
\[c_2 = \frac{1+c_1}{2}\]
or
\[c_2 = \frac{1+c_0}{2}.\]
Then
\[3c_3 - c_2 = 0\]
\[c_3 = \frac{c_2}{3}\]
\[c_3 = \frac{1+c_0}{2\cdot 3}.\]
And so on.

Answer 7

I think you forgot that y' - y was equal to x.

Answer 8

I see - you are right, thank you!