Solving NH 2nd order ODE using Frobenius method

Question

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!

Answer 1

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.

Answer 2

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 $$y=\frac{C_1 e^{\sqrt{2 x}}-C_2 e^{-\sqrt{2 x}}}{\sqrt{2 x}}-x-5$$