I am really stuck with the problem below.
I have the following differential equation: $$x(1+x)y''+ (k + (l + m + 1)x)y'+kly = 0$$ I need to find such functions $\alpha(x)$ and $I(\alpha(x))$ for which $z''+I(x)z=0$, where $z(x)=\frac{y(x)}{\alpha(x)}$.
I first tried to find a particular solution for the first equation in the form of $y=Ax^2+Bx+C$:
$$x(1+x)\times 2A+(k + (l + m + 1)x)\times (2Ax+B)+kl\times (Ax^2+Bx+C) = 0$$
Which transforms to:
$$\bbox[lightgreen, 5px]{2Ax}+\bbox[pink, 4px]{2Ax^2}+\bbox[lightgreen, 5px]{2Akx}+\bbox[pink, 2px]{2A(l+m+1)x^2}+\bbox[lightblue, 5px]{Bk}+\bbox[lightgreen, 3px]{B(l+m+1)x}+\bbox[pink, 5px]{Aklx^2}+\bbox[lightgreen, 5px]{Bklx}+\bbox[lightblue, 5px]{Ckl}=0 $$
Which leads to the following system of equations:
$$ \left\{ \begin{array}{ll} 2A+2A(l+m+1)+Akl&=0 \\ 2A + 2Ak + B(l+m+1)+Bkl &=0 \\ Bk+Ckl&=0 \end{array} \right. $$ And I can't find a solution for this system.
Did I choose the right way to solve the problem? Did I identify the type of the differetial eqiaton correctly? Or should I probably start by first solving $z''+I(x)z=0$?
Any advice would be very much appreciated!