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I've tried but can't get the solution of this ode by Frobenius method.

$(x^2)y''-6y=0$

I tried with $y=\sum_{k=0}^{\infty}(a_k \cdot x^{(k+r)})$ where $a_k$ is coefficient. I can't find the recurrence relation. If any one finds the recurrence relation, that'll do it.

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  • $\begingroup$ Try setting $z=x^2y$, so that $z'=2xy+x^2y'$ and $z''=2y+4xy'+x^2y''$. $\endgroup$
    – user65203
    Jul 10 '16 at 16:43
  • $\begingroup$ My university test paper says to solve it by the method of frobenius. 😞 $\endgroup$
    – mobifz96
    Jul 10 '16 at 16:53
  • $\begingroup$ One doesn't preclude the other. $\endgroup$
    – user65203
    Jul 10 '16 at 16:56
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By Frobenius, the relation between coefficients is

$$x^2k(k-1)a_kx^{k-2}-6a_k=0,$$ i.e. $$a_k=0\lor k(k-1)=6,$$ hence the only nonzero terms are

$$a_3x^3+a_{-2}x^{-2}.$$

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  • $\begingroup$ Is there something missing, in the first line? $\endgroup$
    – mobifz96
    Jul 10 '16 at 16:55
  • $\begingroup$ If you want to do the answer, sir, plz do it with akx^(k+r) $\endgroup$
    – mobifz96
    Jul 10 '16 at 16:57
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    $\begingroup$ @mobifz96: nonsense. $\endgroup$
    – user65203
    Jul 10 '16 at 16:58
  • $\begingroup$ I meant use a solution of this form, not strictly frobenius but tha's what Calcutta University wants us to solve in $y=\sum_{k=0}^{\infty}(a_k \cdot x^{(k+r)})$ $\endgroup$
    – mobifz96
    Jul 12 '16 at 21:11
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    $\begingroup$ @mobifz96: this transformation is trivial, so up to you. $\endgroup$
    – user65203
    Jul 13 '16 at 8:26

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