Question

y'- y = t^2 solution of the equation as a power series??

Answer 1

define y and y' in terms of power series, what do we get?

Answer 2

\[y=\sum_0 c_n~t^n\]
\[y'=\sum_1 c_n~nt^{n-1}\]
use them in the setup

\[\sum_0 c_n~nt^{n-1}~-~\sum_1 c_n~t^n=t^2\]
need to align exponents and indexes

Answer 3

yes

Answer 4

had my 0 and 1 under the sums a bit off track lol

\[\sum_{1} c_n~nt^{n-1}~-~\sum_{0+1} c_{n-1}(n-1)~t^{n-1}=t^2\]
\[\sum_{1} c_n~nt^{n-1}~-~\sum_{1} c_{n-1}(n-1)~t^{n-1}=t^2\]

now we have it so that the indexes and exponents line up, we can combine the summations

Answer 5

\[\sum_{1} [c_n- c_{n-1}(n-1)]~t^{n-1}=t^2\]
lets start poping out terms
\[c_1+\sum_{2} [c_n- c_{n-1}(n-1)]~t^{n-1}=t^2\]
\[c_1+[c_2- c_{1}(1)]~t+\sum_{3} [c_n- c_{n-1}(n-1)]~t^{n-1}=t^2\]
\[c_1+[c_2- c_{1}(1)]~t+[c_3- c_{2}(2)]~t^2+\sum_{4} [c_n- c_{n-1}(n-1)]~t^{n-1}=t^2\]

Answer 6

comparing coefficients:
\[c_1=0\\
c_2-c_1=0\\
c_3-2c_2=1\]
and all the others equal 0

Answer 7

c2=c1=0, c3=1

Answer 8

and for n>3\[c_n- c_{n-1}(n-1)=0\]
\[c_n= (n-1)c_{n-1}\]

c1 = 0
c2 = 0
c3 = 1
c4 = 3 c3 = 3
c5 = 4 c4 = 12
c6 = 5 c5 = 60

etc ...

Answer 9

not real sure how we could develop a rule for the sequence of coeffs tho :/

Answer 10

i spose we could say: y = t^2 + summation formula

Answer 11

any ideas if this has an error in it?

Answer 12

development is c1 + c2.t^2 + c3 t^3...

Answer 13

I think it is correct what you did

Answer 14

c0 + c1 t + c2 t^2 + c3 t^3 etc ....

such that everything but c2 is zero, and c2 = 1

im outta practice to determine if im missing something along the way :/

Answer 15

do we account for the homogenous solution first? then assess it for the particulars?

Answer 16

yes

Answer 17

cn - (n-1)c{n-1} = 0

cn = (n-1)c{n-1}

c1 = 0 c0
c2 = 1*0 c0
c3 = 2*1*0 c0

im gonna need a review :) and i gotta go for tonight