(adsbygoogle = window.adsbygoogle || []).push({}); Question:Find a power series solution in powers of x for the following differential equation

[tex] xy' - 3y = k [/tex]

My attempt:

Assume

[tex] y = \sum_{m=0}^{\infty} a_m x^m [/tex]

So,

[tex] xy' = \sum_{m=0}^{\infty}m a_m x^m [/tex]

[tex] xy'-3y-k=0 [/tex]

implies

[tex] \sum_{m=0}^{\infty}m a_m x^m - 3\sum_{m=0}^{\infty} a_m x^m - k = 0 [/tex]

and

[tex] \left(a_1x+2a_2x^2 +3a_3x^3 +.... \right) - \left(k + 3a_0 + 3a_1x+3a_2x^2+3a_3x^3+... \right) = 0[/tex]

Which means

[tex] a_0=-k/3 [/tex]

[tex]a_1-3a_1=0, a_1=0 [/tex]

[tex]2a_2-3a_2=0, a_2=0 [/tex]

[tex]3a_3 - 3a_3=0, a_3=? [/tex]

[tex]... a_n=0, n>3 [/tex]

The Question:Now, how do I find [tex] a_3 [/tex]?

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# Power series method for DE

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