How do I go about solving this differential equation? $$t^2x''-(6t^4+2t)x'+9t^6x=0$$
I was taught to write as the following $x= t^n+a_1t^{n-1}... \\ x'=nt^{n-1}+a_1(n-1)t^{n-2}...\\ x''=n(n-1)t^{n-2}+a_1(n-1)(n-2)t^{n-3}...$
And then plug those into the differential equation that I have and the coefficient that is associated with $t^n$ would be $0$ and from that equation I would find $n$. Then one of my solutions would be the degree of $n.$ Say $n=1$, then $x=at+b$. From there plug in $x$ into differential equation, from which I could find $a$ and $b$.
Now in this example $n=3$. How is this possible? Could someone do the equation to the end using this method?
 A: Using series, write $$x=\sum_{i=0}^\infty a_it^i$$ $$x'=\sum_{i=0}^\infty ia_it^{i-1}$$ $$x''=\sum_{i=0}^\infty i(i-1)a_it^{i-2}$$ Rewrite $$t^2x''-(6t^4+2t)x'+9t^6x=t^2x''-6t^4x'-2tx'+9t^6x$$which gives $$A=\sum_{i=0}^\infty i(i-1)a_it^{i}-6\sum_{i=0}^\infty ia_it^{i+3}-2\sum_{i=0}^\infty ia_it^{i}+9\sum_{i=0}^\infty a_it^{i+6}=0$$ So, for given power $m$ of $t$, we then have $$m(m-1)a_m-6(m-3)a_{m-3}-2ma_m+9a_{m-6}=0$$ that is to say that the recursion formula is $$m(m-3)a_m=6(m-3)a_{m-3}-9a_{m-6}$$ and we could stop here.
Now, let us look at the beginning of the expansion of $A$; it starts as $$A=-2 a_1 t-2 a_2 t^2+(4 a_4-6 a_1) t^4+(10 a_5-12 a_2) t^5+(9 a_0-18 a_3+18
   a_6) t^6+\cdots$$ so $a_1=0$, $a_2=0$, $a_4=0$, $a_5=0$. So, the arbitrary constants we shall use are $a_0$ and $a_3$. Now, we have all the elements to apply the recursion formula (which will give $a_k=0$ if $k$ is not a multiple of $3$). 
Edit
It is sure that, if you are sufficiently clever (be sure I am not !) to notice that $e^{t^3}$ is a solution (as  Dr. Sonnhard Graubner commented), this makes the problem very simple. Define $x=ye^{t^3}$ and, after simplifcations, the differential equation becomes $$t y''-2 y'=0$$ which is very simple. Reducing the order $p=x'$ and integrating twice let you by the end with $$x(t)=e^{t^3} \left(c_1+c_2 t^3\right)$$
