[tex]

y'' + 4y' + 3y = 0

[/tex]

Firstly, using the characteristic equation I know the general solution is of the form [tex]y\left( x \right) = Ae^{ - x} + Be^{ - 3x} [/tex]. So I know roughly what I should get.

Anyway so if I used series to find the solution to the equation I would try [tex]y = \sum\limits_{n = 0}^\infty {c_n x^n } [/tex]. Then [tex]y' = \sum\limits_{n = 0}^\infty {c_{n + 1} \left( {n + 1} \right)x^n } \Rightarrow y'' = \sum\limits_{n = 0}^\infty {c_{n + 2} } \left( {n + 2} \right)\left( {n + 1} \right)x^n [/tex].

Substituting into the DE: [tex]\sum\limits_{n = 0}^\infty {c_{n + 2} } \left( {n + 2} \right)\left( {n + 1} \right)x^n + 4\sum\limits_{n = 0}^\infty {c_{n + 1} \left( {n + 1} \right)x^n } + 3\sum\limits_{n = 0}^\infty {c_n x^n } = 0[/tex].

[tex]

\Rightarrow \sum\limits_{n = 0}^\infty {\left[ {c_{n + 2} \left( {n + 2} \right)\left( {n + 1} \right) + 4c_{n + 1} \left( {n + 1} \right) + 3c_n } \right]} = 0

[/tex]...This step is only valid provided that the series for y, y', y'' converge in an interval containing x = 0? Anyway...

[tex]

\Rightarrow c_{n + 2} \left( {n + 2} \right)\left( {n + 1} \right) + 4c_{n + 1} \left( {n + 1} \right) + 3c_n = 0

[/tex]

[tex]

c_{n + 2} = \frac{{ - 4c_{n + 1} \left( {n + 1} \right) - 3c_n }}{{\left( {n + 2} \right)\left( {n + 1} \right)}} = - \frac{{4c_{n + 1} }}{{\left( {n + 2} \right)}} - \frac{{3c_n }}{{\left( {n + 2} \right)\left( {n + 1} \right)}}

[/tex]

Ok I'm stuck at this point. I'm actually not really sure what to do. I know that I need an expression for coefficient of y in the original series that I substituted. I'm thinking that I need to solve for c_(n+2). I've seen examples where c_(n+2) is split into c_(2n) and c_(2n+1) so that expressions for the odd and even coefficients are found separately. Is this the general method used to solve second order ODEs using series? Any assistance would be good thanks.