For the following 3 similar Sturm-Liouville problems, I'm struggling to understand how to get to the solutions. I know the solutions already but wolfram alpha gives others, maybe more general ones. The 3 problems are:
(1) $$4y(x)=(x-4)y'(x)+(x^2-16x+55)y''(x)-5x-66, y(0)=26,y'(0)=-15$$ (2) $$4y(x)=(x-6)y'(x)+(x^2-17x+52)y''(x)+4x-52, y(0)=33,y'(0)=-14$$ (3) $$4y(x)=(x-8)y'(x)+(x^2-15x+54)y''(x)+10x-64, y(0)=35,y'(0)=-12$$
The solutions I know are:
(1') $$x^2-15x+26$$ (2') $$x^2-14x+33$$ (3') $$x^2-12x+35$$
Wolfram Alpha's answers may be found in the following:
My question is: How to arrive at the solutions I know, and are wolfram alpha's solutions more general ones?
Taking the first ode,
$$ \left(x^2-16 x+55\right) y''(x)+(x-4) y'(x)-4 y(x)-5 x-66=0 $$
assuming that the solution has the structure
$$ y = a x^2+b x + c $$
after substitution we get
$$ 2 (55 a-2 b-2 c-33)-x (40 a+3 b+5)=0\ \ \ \forall \ \ x $$
then
$$ \cases{ 55 a-2 b-2 c-33=0\\ 40 a+3 b+5=0 }\ \ \ \ \ \ \ \Rightarrow \cases{a = \frac{1}{245}(6c+89)\\ b=-\frac{1}{49}(16c+319)} $$
or
$$ y = \frac{1}{245} (6 c+89) x^2-\frac{1}{49} (16 c+319) x+c $$
and this solution casually admits the two initial conditions but if the conditions were other, the free parameter $c$ at the solution, would not be enough to accept them. I think something similar happens in the other two cases.