<s>[INST] How can I find two independent solution for this ODE? Please help me find two independent solutions for $$3x(x+1)y''+2(x+1)y'+4y=0$$ Thanks from a  new beginner into ODE's.
 [/INST] We suppose that the $\displaystyle y=\sum_{n=0}^{\infty}a_{n}x^n$ is a solution.
Then
$$y=a_{0}+a_{1}x+...+a_{n}x^n+... $$
$$y'=a_{1}+2a_{2}x+3a_{3}x^2+4a_{4}x^3+...+(n+1)a_{n+1}x^n+...$$
$$  y''=2a_{2}+3.2a_{3}x+4.3a_{4}x^2+...+(n+2)(n+1)a_{n+2}x^n+...                $$
replacing this in the DE,
$3x^2y''+3xy''+2xy'+2y'+4y=0$,
we have 
general term
 $$3(n+2)(n+1)a_{n+2}x^{n+2}+3(n+3)(n+2)a_{n+3}x^{n+2}+2(n+2)a_{n+2}x^{n+2}+2(n+3)a_{n+3}x^{n+2}+4a_{n+2}x^{n+2}$$
Then we must have 
$$3(n+2)(n+1)a_{n+2}+3(n+3)(n+2)a_{n+3}+2(n+2)a_{n+2}+2(n+3)a_{n+3}+4a_{n+2}=0$$
or 
$$[3(n+3)(n+2)+2(n+3)]a_{n+3}+[3(n+2)(n+1)+2(n+2)+4]a_{n+2}=0$$
Now you solve this difference equation of the first order to $a_{0}=0$,$a_{1}=1$ and $a_{0}=1, a_{1}=0$.
This give you two solution LI.
Note that $a_{0}$ and $a_{1}$ are the initial condition $y(0)$ and $y'(0)$  
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