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I've been given the problem to solve the following differential equation \begin{equation} x^2y''+(2x+3x^2)y'-2y=0 \end{equation} using Frobenius Method around the regular singular point $x=0$. From the indicialequation I get $r=-2$ or $r=1$, which differ by an integer, giving logaritmic terms in Frobenius method. However, the problem wants me to give two independent power series solutions.

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Let $y=\sum_{n=r}^{\infty}{a_nx^n}$. Then,

$x^2y''+(2x+3x^2)y'-2y=0$ gives us

$\sum_{n=r}^{\infty}{a_nx^n(n(n-1)+2n-2}+a_nx^{n+1}(3n)=0$

$\sum_{n=r}^{\infty}a_nx^n{(n^2+n-2)}=\sum_{n=r}^{\infty}{a_nx^{n+1}(3n)}$

Looking at the $x^r$ term gives us that $r^2+r-2=0$ and $r=-2,1$ as you mentioned.

Comparing the subsequent terms, we get the recursive formula,

$a_n(n^2+n-2)=a_{n-1}3(n-1)$

$a_n=\frac{3}{n+2} a_{n-1}$

So, for $r=1$, we get $a_n=\frac{2\cdot 3^{n}}{(n+2)!}a_1$ And $y=a_1\sum_{n=1}^{\infty}{\frac{2\cdot 3^{n}}{(n+2)!}x^n}$

Do something similar for $r=-2$

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  • $\begingroup$ Yes, I had that but for r=-2 the recursive formula gives you a zero divisor. Variation of constant then gives a logarithmic term but the question specifically asked for power series solutions. $\endgroup$ – ZempZemp Apr 15 '18 at 8:04

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