Local behavior as x→0+ of y'+xy=1/x³

[meenums34]

Asked to find first three terms in the local behavior as x→0+of the solutions of
y′+xy=1/x^3

This was taken by bender and orszag book

Working :
I tried to use method of dominance but later realized that we can find using the series expansion. But I am not sure how to proceed. Please advise how to proceed with it.

 

[HallsofIvy]

Write y as $$\sum a_nx^n$$ where n runs over all integers, both negative and positive. Then $$y'= \sum na_nx^{n-1}$$ and $$xy= \sum a_nx^{n+1}$$.

The equation becomes $$\sum na_nx^{n-1}+ \sum a_nx^{n+1}= x^{-3}$$.
In the first sum let j= n-1 so that we have $$\sum (j+1)a_{j+1}x^j$$.
In the second sum let j= n+ 1 so that we have $$\sum a_{j-1}x^j$$.

Now the equation is $$\sum (j+1)a_{j+1}x^j+ \sum a_{j-1}x^j= \sum ((j+1)a_{j+1}+ a_{j- 1})x^j= x^{-3}$$.

Since power series expansion is unique, we have the (infinite) set of equations
$$(j+1)a_{j+1}+ a_{j-1}= 0$$ for all j except j= -3 and
$$(-3+ 1)a_{-3+ 1}+ a_{-3-1}= -2a_{-2}+ a_{-4}= 1$$.