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For $$x^2y''+x(x^2+1)y'+(x-4)y=0,\tag1$$ there is a regular singular point at $0$, but when I tried to use the Frobenius method and substituted $$y=\sum_{n=0}^\infty a_n x^\left(n+r\right)\tag2$$ into the differential equation, I obtained \begin{multline} (r^2-4)a_0x^r+[(r^2+2r+5)a_1+a_0]x^\left(r+1\right) \\+\sum_{n=0}^\infty\left\{[(n+r+2)^2+2]a_\left(n+2\right)+a_\left(n+1\right)+(n+2)a_n\right\}=0\tag3 \end{multline} I understand that if all $x$ in an interval satisfy the differential equation, then all the coefficients must be zero. But then this means $r=\pm2$ (from the first term), and certainly the second term will not be satisfied by arbitrary $a_0$ and $a_1$ (the initial conditions).

Therefore, my question is does the Frobenius method only work for certain second order linear differential equations with only regular singular points, like where $p(x)$ and $q(x)$ in $~x^2y''+p(x)y'+q(x)y=0~$ are first or second degree polynomials?

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  • $\begingroup$ Frobenius method generally does not allow you to prescribe both $a_0$ and $a_1$. $\endgroup$ – Willie Wong Feb 18 '20 at 19:03
  • $\begingroup$ My understanding is it works in all cases involving only regular singular points. However, regular singular points means there cannot be logarithmic or fractional-power branch points; the singularities in the coefficient ratios must be isolated in the complex domain. Also as noted elsewhere, you cannot in general prescribe both of the first two terms of the power series. The singularity, even when regular, stops you from doing that. $\endgroup$ – Oscar Lanzi Feb 18 '20 at 19:05
  • $\begingroup$ @OscarLanzi Why does a singularity make it impossible to have both $a_0$ and $a _1$ as arbitrary constants? $\endgroup$ – Harry Wang Feb 18 '20 at 21:31
  • $\begingroup$ your equation is missing the $=0$ $\endgroup$ – Aryadeva Feb 18 '20 at 23:04
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You are selecting the values of $r$ specifically so that the first term is zero for any arbitrary value of $a_0$.

You made some sign errors in the collection of coefficients. Using $a_{-2}=a_{-1}=0$, the insertion of the series expansion can be written shortly as $$ \sum_{n=0}^\infty\left\{[(n+r)^2-4]a_n+(n+r-2)a_{n-2}+a_{n-1}\right\}x^{n+r}=0\tag1 $$ Then the first two coefficient identities are $$ [r^2-4]a_0=0\\ [(r+1)^2-4]a_1+a_0=0\tag2 $$ so that still $r=\pm2$, but now $a_1=-\frac1{2r+1}a_0$. Now the question remains if all other coefficients are determined by $a_0$. For $r=2$ this is certainly the case, but for $r=-2$ one gets for $n=1,..,4$ the equation $$ -3a_1+a_0=0\\ -4a_2+a_1-2a_0=0\\ -3a_3+a_2-a_1=0\\ 0a_4+a_3+0a_{2}=0\tag3 $$ which enforces $a_0=0$ so that this branch gives the same solution set as the case $r=2$.

You obtain the second solution by order reduction, as the first solution ($r=2$) has the form $y(x)=x^2a(x)$, $a(x)=\sum_{n=0}^\infty a_nx^n$, $a_0=1$, setting per the reduction-of-order method $y_2=u(x)y(x)=x^2u(x)a(x)$ gives for $u$ the reduced equation \begin{align} y_2'(x)&=u'(x)y(x)+u(x)y'(x)\\ y_2''(x)&=u''(x)y(x)+2u'(x)y'(x)+u(x)y''(x)\\[1em] \hline 0&=x^2[u''(x)y(x)+2u'(x)y'(x)]+x(x^2+1)u'(x)y(x)\tag4\\[1em] \frac{u''(x)}{u'(x)}&=-\frac{2x^2y'(x)+x(x^2+1)y(x)}{x^2y(x)}=-\frac4x-\frac{2a'(x)}{a(x)}-x-\frac1x\tag5 \end{align} so that with the most simple integration constant one finds $$ u'(x)=\frac{e^{-x^2/2}}{x^5a(x)^2}\tag6 $$ While one could directly compute this per power series division and term-wise integration, the structure information alone tells us that $$ u(x)=b_{-4}x^{-4}+...+b^{-1}x^{-1}+\ln(x)+b_1x+...\tag7 $$ so that $$ y_2(x)=x^2\ln(x)a(x)+\sum_{n=0}^\infty c_nx^{n-2}.\tag8 $$


As to the general question, yes, at a regular singularity you get at least one solution in the form of a generalized power series, provided the roots of the indicial equation are real. If they are complex, then with $x^{a+ib}=x^a(\cos(b\ln x)+i\sin(b\ln x))$ one gets some more complicated terms in the real forms of the solution.

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  • $\begingroup$ What is meant by "the first solution has the form $y(x)=x^2a(x)$"? What is "$a(x)$" here? $\endgroup$ – Harry Wang Feb 18 '20 at 21:28
  • $\begingroup$ For $r=2$ you get $y(x)=x^2\sum_{k=0}^\infty a_kx^k$. Taking the special case $a_0=0$ the series is a specific function that can be called $a(x)$. $\endgroup$ – Lutz Lehmann Feb 18 '20 at 21:35
  • $\begingroup$ Would a series solution exist if the differential equation is $x^2y''+xe^x\cos(x)y'+\sin(x)y=0$ instead, since there's still a regular singularity at $0$? $\endgroup$ – Harry Wang Feb 18 '20 at 21:51
  • $\begingroup$ ($a_0=1$ in the previous comment) Yes. Reduced to the lowest degree terms the equation reduces to the Euler-Cauchy form $x^2y''+xy'=0$ with basis solutions $1$ and $\ln x$. So one power series solution $\sum_{n=0}^\infty a_nx^n$, $a_0\ne 0$, will exist. $\endgroup$ – Lutz Lehmann Feb 18 '20 at 22:34
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    $\begingroup$ Yes, of course. One day I will learn to correctly integrate monomials. $\endgroup$ – Lutz Lehmann Feb 18 '20 at 23:24

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