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I'm trying to solve this equation using Frobenius method. $$ xy''-y'-4x^3y=0 $$


$$ y=\sum a_n x^{n+r} $$ However doing the indicial equations I get an incompatible system. $$ (r(r-1)-r)a_0x^{r-2}=0 \\ (r+1)(r-1))a_1x^{r-1}=0 $$

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    $\begingroup$ Why do you think that you get two indicial equations? For $a_0\ne 0$ (why else would you start the power series at index $0$) you get $r(r-2)=0$, so that you have a solution for $r=0$ and possibly another independent one for $r=2$. It might also be that the second solution requires order reduction and contains some logarithmic term. $\endgroup$ – Lutz Lehmann Oct 19 '19 at 17:56
  • $\begingroup$ But what happens with the other equation? If these terms aren't 0 I cant equal the series one to 0 $\endgroup$ – Marco Villalobos Oct 19 '19 at 18:28
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    $\begingroup$ Of course you can satisfy the second equation, just set $a_1=0$. $\endgroup$ – Lutz Lehmann Oct 19 '19 at 18:52
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For $r=0$ you get the coefficient iteration $$ n(n+1)a_{n+1}-(n+1)a_{n+1}-4a_{n-3}=0 $$ where it is understood that $a_k=0$ for $k<0$. This iteration formula tells us that the power series coefficients split into 4 independent sub-sequences $(a_{4k+i})_k$, $i=0,1,2,3$. Only the ones with $i=0$ and $i=2$ are non-zero. The second series for $i=2$ also accounts for the case $r=2$. As these sub-sequences give independent solutions, this is a full generating system, a solution basis.


If you set, inspired by the coefficient structure, $y(x)=f(x^2/2)$, then $$ 0=x[x^2f''(x^2/2)+f'(x^2/2)]-xf'(x^2/2)-4x^3f(x^2/2)=x^3(f''(x^2/2)-4f(x^2/2)] $$ which implies that $f(t)=c_1e^{2t}+c_2e^{-2t}=d_1\cosh(2t)+d_2\sinh(2t)$, so that $$ y(x)=a_0\cosh(x^2)+a_1\sinh(x^2). $$

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  • $\begingroup$ And why does the formula tell us there are 4 independent subsequences $\endgroup$ – Marco Villalobos Oct 19 '19 at 18:41
  • $\begingroup$ How i solve one of this iterations $\endgroup$ – Marco Villalobos Oct 19 '19 at 18:46
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    $\begingroup$ The formula collects the coefficients of $x^n$. There are 4 subsequences because $n(n-2)a_{n}=4a_{n-4}$ connects only the elements of $(a_{4k+i})_k$. And what do you mean with "solve"? What you can do is to set $y(x)=u(x^4)$ or $y(x)=x^2v(x^4)$ and try to get something close to a Bessel equation. $\endgroup$ – Lutz Lehmann Oct 19 '19 at 18:56
  • $\begingroup$ I think a solution is cosh(x^2) how can i obtain it from this recursssion $\endgroup$ – Marco Villalobos Oct 19 '19 at 19:02
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    $\begingroup$ You get $a_{4k+4}=\frac{4}{(4k+4)(4k+2)}a_{4k}=\frac1{(2k+2)(2k+1)}a_{4k}$, so that $(2(k+1))!a_{4(k+1)}=(2k)!a_{4k}$ which gives $a_{4k}=\frac{a_0}{(2k)!}$ which are the coefficients of the hyperbolic cosine. $\endgroup$ – Lutz Lehmann Oct 19 '19 at 19:06

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