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Having the equation $$x^{2}y''+xy'+x^{2}y=0$$

I get the indicial equation at get r=0, and am left with the equation. $$r^{2}a_{0}x^{r}+(r^{2}+2r+1)a_{1}x^{r+1}+\sum^{\infty}_{0}\big[[(n+r+2)(n+r+1)+(n+r+2)]a_{n+2}+a_{n}\big]x^{n+r+2}=0$$

subbing in r=0 gives: $$a_{1}x+\sum^{\infty}_{0} [(n+2)^2a_{n+2}+ a_{n}]x^{n+2}=0$$

So im not sure how to get the recurrence relation as i don't have an $a_{n+1}$ to find this and also i have an $a_{1}$ term so am not sure how to find this. Any help would be appreciated thank you.

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Note that the second term in your second equation implies that $a_1 = 0$ and thus $$a_1 = a_3 = a_5 = \cdots = 0 $$ by the recurrence relation. Now compute $$a_2, a_4, a_6 \dots$$ using the recurrence relation. A clear pattern should emerge that will allow you to write down an expression for $a_{2k}$. Then, a solution is $$y=\sum_{n=0}^\infty a_{2n} x^{2n}.$$

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  • $\begingroup$ This is flat-out wrong. Even though all coefficients go to zero at $x=0$, the ratio $Q/P$ is singular there making $x=0$ a singular point. $\endgroup$
    – Oscar Lanzi
    Apr 25, 2016 at 2:21
  • $\begingroup$ @OscarLanzi Are you sure? I was taught in my PDE's class that at least one of $Q$ or $R$ is non-zero to get a RSP. Feel free to edit my answer if you are absolutely sure. I certainly don't want to mislead the OP. $\endgroup$
    – MathMajor
    Apr 25, 2016 at 2:25
  • $\begingroup$ This is Bessel's equation and the solution is $y=c_1J_0(x)+c_2N_0(x)$ and $N_0(x)$ does indeed have a logarithmic singularity at the origin. So some blood will be shed on the way to the general solution. $\endgroup$
    – user5713492
    Apr 25, 2016 at 3:25

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