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Quite stumped with this one so far. I have the following non-homogeneous ODE:

$$2x^2y''+3xy'-xy=x^2+2x$$

And I need to find a solution for $x_0<0$ using Frobenius. Obviously we can center the solution about $x_0=0$, and we can see it is a regular singular point based on the coefficients of $y'$ and $y$. But I searched quite a bit and couldn't find anything about the non-homogeneous part of the equation (every example that I find on here either is homogeneous, or the NH part is a constant). I know for $p(x)$ and $q(x)$ we check for RSP such that:

$$xp=x\left(\frac{3x}{2x^2}\right)=3/2$$ $$x^2q=x^2\left(\frac{-x}{2x^2}\right)=-x/2$$

Which are analytical for $x_0=0$, so no problem there. But I have no clue what to do with the non-homogeneous part of the equation to verify that it is really a RSP. Of course the problem would easily be solved simply with a regular power series, but I absolutely have to use Frobenius without using series to check for analyticity. Any pointers?

Apologies if the terminology is a bit off, english is not my mother tongue. Thanks in advance!

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  • $\begingroup$ @Raffaele In the normal notation of Frobenius for second order equations, the leading coefficient is made to be $x^2$; if it isn't already $x^2$, you force it to be, even if it superficially makes the equation more complicated. $\endgroup$
    – Ian
    Feb 10, 2021 at 18:26
  • $\begingroup$ I don't know if I understood well. Are you looking for the particular solution? $\endgroup$
    – Raffaele
    Feb 10, 2021 at 18:38
  • $\begingroup$ @Raffaele Seems like they want the general solution, and know how to get the homogeneous solution, so it suffices to find a particular solution. $\endgroup$
    – Ian
    Feb 10, 2021 at 18:59
  • $\begingroup$ Yes, I am looking for the particular solution indeed. Apologies if it isn't clear, as I said I'm trying to translate terms the best I can. The other answers are very useful, but since this is for an assignment I am forced to use Frobenius method, so I can't really use the method with the polynomial. The way we saw it is that we find the singular point upon which we want to expand the solution (in this case 0). We then divide each coefficient by the 2nd order coefficient, then multiply the first order coefficient by $(x-x_0)$ and the 0th(?) order coefficient by $(x-x_0)^2$ (Next post) $\endgroup$
    – Jordan
    Feb 10, 2021 at 19:35
  • $\begingroup$ If both functions are analytical, $x_0$ is a regular singular point and we can use it with Frobenius method. This works perfectly well for homegeneous equations or equations where the NH side is a constant, but I can't find any information about a non-constant NH equation. I have no idea if this method is standard, but I am kinda forced into using it. $\endgroup$
    – Jordan
    Feb 10, 2021 at 19:37

2 Answers 2

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Suppose that the particular solution is a polynomial $$y_p=ax^2+bx+c$$ We have $$y'=2ax+b;\;y''=2a$$ so the DE becomes $$4ax^2+3 x (2 a x+b)-x \left(a x^2+b x+c\right)\equiv x^2+2 x$$ that is $$-a x^3+x^2 (10 a-b)+x (3 b-c)\equiv x^2+2x$$ this means $$ \begin{cases} -a=0\\ 10a-b=1\\ 3b-c=2\\ \end{cases} $$ which gives $a=0;b=-1;c=-5$ so the particular solution is $y_p=-x-5$

Added to the general solution we have $$y=\frac{C_1 e^{\sqrt{2 x}}-C_2 e^{-\sqrt{2 x}}}{\sqrt{2 x}}-x-5$$

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It would be convenient if there was just a low degree polynomial solution, so let's see if there is by plugging in a few monomials into the LHS:

$$L(1)=-x \\ L(x)=3x-x^2.$$

This is already enough play to get what you want:

$$L(bx+a)=-ax+3bx-bx^2$$

and so you want

$$-b=1 \\ -a+3b=2.$$

Note that this would have failed if the RHS had a constant term.

Now $z=y-(bx+a)$ solves the homogeneous equation, and $bx+a$ is analytic, so the analytical behavior of $y$ and $z$ is the same, after adjusting the initial data appropriately.

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  • $\begingroup$ Thank you for the answer. Although the regular power series solution would be appropriate here, I absolutely have to solve it using Frobenius method, so I can't really resort to using that unfortunately. $\endgroup$
    – Jordan
    Feb 10, 2021 at 18:09
  • $\begingroup$ @Jordan It's actually even easier to find a single inhomogeneous solution than I was saying. I've edited the answer. $\endgroup$
    – Ian
    Feb 10, 2021 at 18:20
  • $\begingroup$ any idea why this method fails when RHS is constant term, say 1? any suggestion what to do if RHS is constant term then? $\endgroup$
    – Nasser
    Oct 7, 2021 at 9:24
  • $\begingroup$ @Nasser Then you would have to have a singularity at $x=0$ to achieve that. $\endgroup$
    – Ian
    Oct 7, 2021 at 12:54

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