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I'm trying to find two independent series solutions, expanded about x = 0, that satisfy:

$$ g''+2xg'+4g=0 $$

so far I have gotten the indicial equation and found $r=0$ and $r=1$. Then I substitute the derivatives by the respective sigma notation and found $a_n=\frac{-2}{n+r-1}$.

Now, if $r=0$ then $a_n=\frac{-2}{n-1}$.

And if $r=1$ then $a_n=\frac{-2}{n}$.

I tried from here to collect coefficients and find the respective power series but I do not understand how there 2 independent series solutions.

Also, on the problem it's given the general solution that I'm trying to approach which is:

$$g(x)=Axe^{-x^2}+B\Sigma^{\infty}_{n=0}\frac{(-4)^nn!}{(2n)!}x^{2n}$$ with A and B arbitrary constants.

Any suggestions?

Thanks in advance!

Aline Bellangero
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  • What you have to do is the following. Suppose that you have a power series $\sum a_n x^n$ solution of the equation, of radius of convergence $>0$. Then you can differentiate it term by term, so inject $\sum a_n x^n$ in the equation. Do some manipulation to rassemble all the same powers together, until you reach an equation $\sum b_n x^n = 0$. This implies that all the $b_n$'s are zero, so you will get some conditions on the $a_n$'s (probably recursive conditions). Now, choosing different initial values for $a_0$ and $a_1$ will lead you to independant solutions of the equation. – TheSilverDoe Nov 05 '20 at 07:53

1 Answers1

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If you plug $\sum_{n=0}^\infty x^n$ in your equation we get $$ \sum_{n=2} a_n n (n-1)x^{n-2}+2\sum_{n=1}^\infty a_n n x^n +\sum_{n=0} 4 a_n x^n=0. $$ The left-hand side can be written as a unique series $$ \sum_{n=0}^\infty \underbrace{[a_{n+2} (n+2)(n+1)+a_n(2n+4)}_{=:b_n} x^n=0. $$ Hence we need all the coefficients $b_n =0$ and this gives a recurrent relation $f(a_n)=a_{n+2}$. If we choose $a_1$ we find all the odd terms ($a_{2k+1}$) and if we choose $a_2$ we determine all the even terms $a_{2k}$.

You can look here to understand why these solutions are independent.

If you Taylor expand the general solution given by the text you get your 2 solutions (observe that $xe^{-x^2}$ has only odd terms in the expansion and the second one only the odd ones, hence $A,B$ corresponds to a rescale of $a_1,a_0$ respectively).

Jonas
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