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As the title says, I have the equation $x^4y''=y$ And then assume a solution of $\sum a_n x^{n+k}$ This gives the follwing: $$\sum a_n(n+k)(n+k-1)x^{n+k-2}=\sum a_n x^{n+k-4}$$ I can expand the first few terms as $$a_o\,k(k-1)x^{k-2}+a_1\, k(k+1)+...-a_0 x^{k-4}-a_1x^{k-3}-...=0$$ The problem says to assume $a_0\neq 0$, so I equate like terms and I immediately get that $a_0$=0.

Does this mean that no solutions exist in the form of a frobenius series?

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As you say, the relation forces $a_0=a_1=0$. And then the recursion will give you $a_n=0$ for all $n $, rendering the solution $y (x)=0$. No other solution has the form required by Frobenius method.

(According to Wolfram Alpha, a fundamental set of solutions is given by $xe^{1/x} $ and $xe^{-1/x} $).

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  • $\begingroup$ It seems to me that $y=c_1 \,x\,e^{1/x} +c_2\, x\, e^{-1/x}$ $\endgroup$ – Claude Leibovici Nov 10 '15 at 6:42
  • $\begingroup$ Yes, I missed the first $x $. Thanks. $\endgroup$ – Martin Argerami Nov 10 '15 at 12:52

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