I am trying to solve the problem using the Frobenius method around x=0 $$3x^2y''-xy'+y=0$$ $$F(r)=r(r-1)+p_0r+q_0$$ since $$p(x)=-1/3$$ $$q(x)=1/3$$ then $$F(r)=r^2-4/3r-1/3=0$$ $$r=2/3±\sqrt{7}/3$$ the difference between r1 and r2 is a non integer so we use the following method of substitution $$y(x)=x^r\sum_{n=0}^\infty a_nx^n = \sum_{n=0}^\infty a_nx^{n+r}$$ we substitute y(x) into the equation $$ 3x^2{\sum_{n=0}^\infty {(n+r+1)(n+r+2)}a_{n+2}x^{n+r}} - x{\sum_{n=0}^\infty (n+r+1)a_{n+1}x^{n+r}}+{\sum_{n=0}^\infty a_nx^{n+r}} = 0$$ then $$ 3{\sum_{n=2}^\infty {(n+r-1)(n+r)}a_{n}x^{n+r}} - {\sum_{n=1}^\infty (n+r)a_{n}x^{n+r}}+{\sum_{n=0}^\infty a_nx^{n+r}} = 0$$ which gives us the relationship: $$(3(n+r-1)(n+r)-(n+r)+1)a_n = 0\,\,\,\,\,\,\,\,\,\,\,n\ge2$$ $$(-(n+r)+1)a_n = 0\,\,\,\,\,\,\,\,\,\,\,n=1$$ $$a_n = 0\,\,\,\,\,\,\,\,\,\,\,n=0$$
For all n a is zero. I don't know where I'm making a mistake, there should be a solution. Any help would be appreciated.
This ODE does not require the method of Frobenius. Notice that the degree of the monomial coefficient matches the order of the derivative it is attached to. This means we will have two monomial homogeneous solutions of the form $y=x^r$:
$$y = x^r \implies (3r^2-4r+1)x^r = (3r-1)(r-1)x^r = 0$$
therefore
$$y = C_1 x^{\frac{1}{3}}+C_2x$$