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I've seen a lot of answers online about how to solve these but they're all constant coefficients

Find the power series of: $$y'' + xy' + y = x^2 + 2x + 1 $$in powers of $x$ (given that $x_0=0$)

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  • $\begingroup$ Can you check that you have rewritten the problem correct? $\endgroup$ Apr 10, 2020 at 6:51
  • $\begingroup$ Yes, that's indeed the problem, I've triple checked $\endgroup$ Apr 10, 2020 at 7:11
  • $\begingroup$ What is $x_0$ ? $\endgroup$
    – user65203
    Apr 10, 2020 at 9:26

1 Answer 1

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Let $y(x)=\sum_{n\geq 0}a_nx^n$, then

  • $xy'(x)=\sum_{n\geq 0}na_nx^n$
  • $y''(x)=\sum_{n\geq 0} (n+2)(n+1)a_{n+2}x^n$

Now consider $y''(x)+xy'(x)+y(x)=0$, which happens when $$\sum_{n\geq 0}[(n+2)(n+1)a_{n+2}+(n+1)a_n]x^n\tag{1}$$ vanishes for all $x$. Thus $(n+2)(n+1)a_{n+2}+(n+1)a_n=0$, which simplifies to $$a_{n+2}=-\frac{a_n}{n+2}\tag{2}$$ If $a_0$ and $a_1$ are given, then all $a_n$ can be determined by this recursion formula. The values of $a_0$ and $a_1$ can be determined by $y(0)$ and $y'(0)$.

For a particular solution of the non-homogeneous equation, take $y_p(x)=c_0+c_1x+c_2x^2$ and insert this into the given equation, which is $$3c_2x^2+2c_1x+2c_2+c_0=x^2+2x+1$$ and then you will find

  • $3c_2=1$
  • $2c_1=2$
  • $2c_2+c_0=1$

by comparision. This will result in $y_p(x)=\frac{1}{3}x^2+x+\frac{1}{3}$.

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