Ordinary DEQ - Frobenius Method I have a DEQ:
$$y''-\frac{6}{x}y'+\frac{12+x^2}{x^2}y=0$$
Then in series form:
$$\sum_{m=0}^{\infty}(m+r)(m+r-1)a_mx^{m+r-2}-6\sum_{m=0}^{\infty}(m+r)a_mx^{m+r-2}+12\sum_{m=0}^{\infty}a_mx^{m+r-2}+\sum_{m=0}^{\infty}a_mx^{m+r}$$
The lowerst power is obviously $m+r-2$, and we get the indicial equation $r(r-1)-6r=0 \rightarrow r_1=0, r_1=7$
Next, we equate the powers, since the first 3 are the same, we change the index on the 4th series:
$$s_4=\sum_{m=2}^{\infty}a_{m-2}x^{m+r-2}$$
Rearranging the series:
$$((m+r)(m+r-1)-6(m+r)+12)a_m+a_{m-2}=0$$
Let $r=0$, i get:
$$a_m=-\frac{a_{m-2}}{m(m-7)+12}$$
Then $$a_2=-\frac{a_0}{30}, a_4=-\frac{a_2}{56}=-\frac{a_0}{1680}.....$$
now that just doesn't seem very likely,,i triple-checked my solution, and don't know where i might have made a mistake, can someone have a look?

As of @JerryHolmes' request, I will show how I solved it.
There is no need to overcomplicate stuff by using Frobenius' method; we can seek a simple power-series solution
$$y(x)=\sum_{n=0}^\infty a_nx^n.$$
If we substitute the series with its derivatives into ODE, you should get
$$\sum_{n=0}^\infty(n-4)(n-3)a_nx^{n-2}+\sum_{n=0}^\infty a_nx^n=0.$$
Notice that to combine the $2$ series, we have to set the first $2$ terms of the lagging series equal to $0.$ If we do so, we get $a_0=a_1=0.$
Now we can combine the $2$ series if we increase the index of the lagging series by $2$
$$\sum_{n=0}^\infty((n-1)(n-2)a_{n+2}+a_n)x^n=0.$$
So we get
$$(n-2)(n-1)a_{n+2}+a_n=0.\tag{$\star$}\label{1}$$
Notice that we are going to solve for the coefficients via the $a_{n+2}$ term, so we don't want its coefficient in $\eqref{1}$ to be $0$ because if that happens, we can't solve for them so we will have to assume them as initial conditions. This happens for $n=1,2\implies a_{n+2}=a_3,a_4.$
So we know (or delt with) $a_0,a_1,a_3,a_4.$ To solve for $a_2,$ plug $n=0$ into $\eqref{1}$ to get $a_2=0.$
Now, we need to solve for $a_{n>4}.$ We can now safely solve explicitly for $a_{n+2}$ without the risk of dividing by $0.$
$$a_{n+2}=-\frac{a_n}{(n-2)(n-1)}.$$
Since the coefficients jump by $2,$ an even coefficient will depend on the previous even one and the same follows for the odd coefficients.
$$a_{2n+1}=-\frac{a_{2n-1}}{(2n-3)(2n-2)}\\
a_{2n}=-\frac{a_{2n-2}}{(2n-4)(2n-3)}$$
To avoid confusion in this part, set $2k+1=n+2$ to solve for the odd coefficients and set $2k=n+2$ to solve for the even coefficients.
Since we start with $a_3$ and $a_4,$ odd coefficients will depend on $a_3$ and even coefficients will depend on $a_4.$
If you solve for the first few coefficients (or just notice the fact that we just keep dividing by consecutive integers,) you will notice the pattern
$$a_{2n+1}=(-1)^{n+1}\frac{a_3}{(2n-2)!}\\
a_{2n}=(-1)^{n+1}\frac{a_4}{(2n-3)!}.$$
Now we can write the general solution
$$y(x)=a_3\sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n-2)!}x^{2n+1}+\\a_4\sum_{n=2}^\infty\frac{(-1)^{n+1}}{(2n-3)!}x^{2n}.$$
If we notice that $a_3=\frac{y^{(3)}(0)}{3!}$ and $a_4=\frac{y^{(4)}(0)}{4!}$ because of Mclaurin series, we can write it as
$$y(x)=\frac{y^{(3)}(0)}{3!}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n-2)!}x^{2n+1}+\\\frac{y^{(4)}(0)}{4!}\sum_{n=2}^\infty\frac{(-1)^{n+1}}{(2n-3)!}x^{2n}.$$
We can factor $x^3$ and increase the index of the first series by $1$ and that of the second series by $2$ to get
$$y(x)=x^3\left(\frac{y^{(3)}(0)}{3!}\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}-\\\frac{y^{(4)}(0)}{4!}\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}x^{2n+1}\right)$$
which can be written in terms of familiar functions
$$y(x)=x^3\left(\frac{y^{(3)}(0)}{3!}\cos x-\\\frac{y^{(4)}(0)}{4!}\sin x\right)$$
as your professor said.
