Consider the ODE $$x^2 y''(x) + 6x y'(x) - 9y(x) =0 .$$
It is clear that we can solve the ODE by the method of reduction of order. However, if we "see" the function $y$ as some constant just for a second, we can easily identify this equation as an algebraic equation similar to $$(x-3)^2 = 0.$$
Question:
Is there any method for solving this kind of ODEs that are similar to solving algebraic equations as above ?
Motivation:
I mean for a given $x = x_0$, the function will have some fix value, so (bear with me for a second) if we plug $x_0$ into $y(x)$ without putting the value $x_0$ to $x$ in the coef. of the ODE (we are going to plug it into, but see this as there is some delay in the process :) ), we can treat the equation as an algebraic equation, and this can be done for any values of $x$, and this kind of thouhgt lead me to above question about whether such a thing can be justified rigorously ?
Your equation is not equivalent to $$ (xD-3)^2y=0 $$ as $x$ and $D=\frac {d}{dx}$ do not commute. Indeed you get $Dxf(x)=f(x)+xf'(x)=(1+xD)f(x)$ so that $$ (xD-3)^2=xDxD-6xD+9=x(1+xD)D-6xD+9=x^2D^2-5xD+9. $$
In reverse it means that you would have to factorize $$ x^2D^2-6xD+9=(xD)^2-7xD+9=(xD-3.5)^2-3.25. $$ or in the given equation $$ x^2D^2+6xD-9=(xD)^2+5xD-9=(xD+2.5)^2-15.25. $$
Once you have this factorization $(xD-a)(xD-b)y=0$ you can form an adapted first order system $$ (xD-b)y=v\\ (xD-a)v=0 $$ where you usually get powers of $x$ and their linear combinations as solutions.